th
5 Year
Applied Maths
Higher Level
Kieran Mills
Uniform Accelerated Motion
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Contents: Uniform Accelerated Motion
Section 1 Using the Formulae....................................................................2
Exercise 1...................................................................................6
Exercise 2...................................................................................8
Section 2 Successive Times and Distances.................................................10
Exercise 3...................................................................................14
Section 3 Catch-up Problems.....................................................................16
Exercise 4...................................................................................25
Section 4 Velocity-time Graphs.................................................................28
Exercise 5...................................................................................39
Section 5 Free-fall.....................................................................................42
Exercise 6...................................................................................46
Leaving Cert Questions 2014-1996..............................................................49
©The Dublin School of Grinds
Page 1
Kieran Mills & Tony Kelly
©The Dublin School of Grinds
Page 2
Kieran Mills & Tony Kelly
A
u = 10 m s-1
Quantity
SymbolUnits
Acceleration
a
m s–2
Initial velocity
u
m s–1
Final velocity
v
m s–1
Time
t
s
Displacement
sm
0m
a = 4 m s-2
t = 10 s
Section 1: Using the Formulae
+
Right
B
v = 50 m s-1
s = vt − 12 at 2 ....(5) [ No u ]
v 2 = u 2 + 2as...(4) [ No t ]
s = ut + 12 at 2 ....(3) [ No v]
t =?
v = 50 m s −1
u = 10 m s −1
a = 4 m s −2
v = u + at
50 = 10 + 4t
40 = 4t
∴ t = 10 s
= 300 m
= 12 (60)(10)
= 12 (50 + 10)(10)
s = 12 (v + u )t
Ex. How long does it take the bus to travel from A to B?
How far is it from A to B?
Solution
s = 300 m
+a: acceleration
–a: deceleration
-
Left
s = 12 (v + u )t....(2) [ No a ]
v = u + at.........(1) [ No s ]
Equations of Motion
©The Dublin School of Grinds
Page 3
Kieran Mills & Tony Kelly
Example 6
s = 8 m, u = 6 m s–1, a = –2 m s–2, t = ?
Example 5
s = 120 m, v = 10 m s–1, t = 3 s, a = ?
Example 4
u = 4 m s–1, v = 9 m s–1, a = 0.5 m s–2, s = ?
Example 3
u = 8 m s–1, v = 15 m s–1, t = 6 s, s = ?
Example 2
u = 6 m s–1, a = 10 m s–2, t = 4 s, v = ?
Example 1
u = 3 m s–1, a = 6 m s–2, t = 4 s, s = ?
©The Dublin School of Grinds
Page 4
Kieran Mills & Tony Kelly
= 12 + 48
= 60 m
= (3)(4) + 12 (6)(4) 2
= 46 m s
−1
= 3(23)
= 69 m
= 12 (15 + 8)6
s = 12 (v + u )t
No a present. Use Equation (2).
No s present. Use Equation (1).
v = u + at
= 6 + 10(4)
= 6 + 40
No v present. Use Equation (3).
s = ut + 12 at 2
u = 8 m s–1
v = 15 m s–1
t=6s
s=?
u = 6 m s–1
a = 10 m s–2
t=4s
v=?
Mathematical Calculations
Mathematical Calculations
Mathematical Calculations
u = 3 m s–1
a = 6 m s–2
t=4s
s=?
Example 3
u = 8 m s–1, v = 15 m s–1, t = 6 s, s = ?
Solution
Example 2
u = 6 m s–1, a = 10 m s–2, t = 4 s, v = ?
Solution
Example 1
u = 3 m s–1, a = 6 m s–2, t = 4 s, s = ?
Solution
©The Dublin School of Grinds
Page 5
Kieran Mills & Tony Kelly
92 = 42 + 2(0.5) s
81 = 16 + s
s = 65 m
v 2 = u 2 + 2as
No t present. Use Equation (4).
u = 4 m s–1
v = 9 m s–1
a = 0.5 m s–2
s=?
Mathematical Calculations
Example 4
u = 4 m s–1, v = 9 m s–1,
a = 0.5 m s–2, s = ?
Solution
a = −20 m s −2
240 = 60 − 9a
9a = −180
120 = 30 − 92 a
120 = 10(3) − 12 a (3) 2
s = vt − 12 at 2
No u present. Use Equation (5).
s = 120 m
v = 10 m s–2
t=3s
a=?
Mathematical Calculations
Example 5
s = 120 m, v = 10 m s–1, t = 3 s, a = ?
Solution
t 2 − 6t + 8 = 0
(t − 4)(t − 2) = 0
t = 2 s, 4 s
8 = 6t − t 2
8 = (6)t + 12 (−2)t 2
s = ut + 12 at 2
No v present. Use Equation (3).
s=8m
u = 6 m s–1
a = –2 m s–2
t=?
Mathematical Calculations
Example 6
s = 8 m, u = 6 m s–1, a = –2 m s–2,
t=?
Solution
Exercise 1. Using the formulae
1.
u = 0 m s −1 , a = 4 m s −2 , t = 6 s, s = ?
2.
u = 5 m s −1 , a = 3 m s −2 , t = 3 s, v = ?
3.
u = 0 m s −1 , a = 4 m s −2 , t = 4 s, s = ?
4.
u = 7 m s −1 , v = 15 m s −1 , t = 10 s, s = ?
5.
u = 4 m s −1 , v = 10 m s −1 , s = 7 m, a = ?
6.
v = 30 m s −1 , a = 6 m s −2 , t = 2 s, u = ?
7.
u = 6 m s −1 , v = 3 m s −1 , s = 1.5 m, a = ?
8.
s = 35 m, a = 6 m s −2 , t = 2 s, u = ?
9.
s = 40 m, u = 5 m s −1 , v = 15 m s −1 , t = ?
−1
−1
−2
10. u = 5 m s , v = 10 m s , a = 0.5 m s , s = ?
−1
−1
11. u = 48 m s , v = 12 m s , t = 9 s, a = ?
−2
12. s = 30 m, t = 2 s, a = 4 m s , v = ?
−1
13. s = 120 m, t = 3 s, v = 10 m s , a = ?
−2
−1
14. s = −4 m, a = 2 m s , v = 3 m s , t = ?
−1
−2
15. s = 8 m, v = 7 m s , a = 3 m s , t = ?
−1
−2
16. s = 24 m, u = 5 m s , a = 2 m s , t = ?
−1
−2
17. s = 8 m, u = 6 m s , a = −2 m s , t = ?
−1
−2
18. v = 72 km h , a = 0.5 m s , t = 1 min, u = ?
−1
−2
19. s = 8 m, v = 8 m s , a = 3 m s , u = ?
−1
−2
20. s = −6 m, v = 4 m s , a = 4 m s , t = ?
©The Dublin School of Grinds
Page 6
Kieran Mills & Tony Kelly
Answers
Exercise 1
1. 72 m
11. –4 m s–2
2. 14 m s–1
12. 19 m s–1
3. 32 m
13. –20 m s–2
4. 110 m
14. 4 s
5. 6 m s–2
15. 2 s,
6. 18 m s–1
16. 3 s
7. –9 m s–2
17. 2s, 4s
8. 11.5 m s–1
18. –10 m s–1
9. 4 s
19. 4 m s–1
10. 75 m
20. 3 s
©The Dublin School of Grinds
Page 7
8
3
s
Kieran Mills & Tony Kelly
Exercise 2. Simple problems using the formulae
1.
A train starts from rest and accelerates uniformly at 2.5 m s–2 until it reaches a speed of
25 m s–1. Find the distance moved and the time taken for this motion.
2.
A car can accelerate from rest to 90 km h–1 in 7.5 seconds. Find its acceleration.
3.
In travelling 65 cm along the barrel of a rifle a bullet accelerates from rest to 230 m s–1.
Find the accceleration and the time the bullet is in the barrel.
4.
A car travelling at 24 m s–1 requires a minimum braking distance of 36 m. What is its
deceleration? How long does it take to stop?
5.
A car starts from rest with acceleration 4 m s–2. How far does it go in (i) 2 s, (ii) 3 s,
(iii) the third second.
6.
A body moves in a straight line and increases its velocity from 3 m s–1 to 15 m s–1
uniformly in 6 s. Find the acceleration and the distance travelled.
7.
A particle starts with a velocity of 3 m s–1 and accelerates uniformly at 1.5 m s–2.
How far does it go in (i) 1 s, (ii) 5 s, (iii) the fourth second.
8.
A body is projected from the origin with a velocity of 8 m s–1 and acceleration –2 m s–2.
Find
(i) the velocity when t = 3 s,
(ii) when it comes to instantaneous rest.
9.
A particle moves along a straight line between two points P and Q with constant
acceleration 0.8 m s–2. Its velocity at Q is 1.2 m s–1 greater than the velocity at P.
If the distance PQ is 48 m, find the velocity at P.
How long after passing P does it take the velocity to reach 48 m s–1.
10.
A car is moving with speed u m s–1. The brakes of the car can produce a constant
2
deceleration of 5 m s–2. It is known that when the driver decides to stop, a period of 5 s
elapses before the brakes are applied. As the car passes a point O, the driver decides to
stop.
Find in terms of u the minimum distance of the car from O when the car comes to rest.
The driver is approaching traffic lights and is 102 m away when the light changes from
green to amber. The lights remain amber for 3 s before changing to red.
Show
(a) when u < 30 the driver can stop before reaching the lights,
(b) when u > 34 the driver can pass the light before it turns red.
©The Dublin School of Grinds
Page 8
Kieran Mills & Tony Kelly
Answers
Exercise 2
1. 125 m, 10 s
2.
10
3
m s −2
3. 40, 692 m s–1, 5.652 ×10−3 s
4. –8 m s–2, 3 s
5. 8 m, 18 m, 10 m
6. 2 m s–2, 54 m
7. 3.75 m, 33.75 m, 8.25 m
8. 2 m s–1, 4 s
9. 31.4 m s–1, 20.75 s
10.
1
10
u 2 + 52 u
©The Dublin School of Grinds
Page 9
Kieran Mills & Tony Kelly
©The Dublin School of Grinds
Page 10
Kieran Mills & Tony Kelly
A uniformly decelerating body covers successive 100 m distances in 5 s
and 10 s. Find its initial speed, the deceleration and the further time for
the body to come to rest.
Example 2
In two successive seconds a uniformly accelerating body travels 5 m and
13 m. Find its acceleration and its initial velocity.
Example 1
Section 2: Successive Times and Distances
©The Dublin School of Grinds
Page 11
Kieran Mills & Tony Kelly
u
s = 13 m
s=5m
18 = 2u + 2a....(2)
18 = u (2) + 12 a (2) 2
t = 2 s, s = 18 m
t=1s
t=1s
a
10 = 2u + a....(1)
5 = u + 12 a
5 = u (1) + 12 a (1)
2
s = ut + 12 at 2
In these types of problems always use the equation shown.
Take all values from the beginning point.
Answers
a = 8 m s −2 , u = 1 m s −1
10 = 2u + a....(1)
10 = 2u + 8
2 = 2u
∴u = 1
−10 = −2u − a
18 = 2u + 2a
8=a
10 = 2u + a......(1)(× − 1)
18 = 2u + 2a....(2)
Mathematical Calculations
Example 1
In two successive seconds a uniformly accelerating body travels 5 m and 13 m. Find its acceleration and its initial
velocity.
Solution
©The Dublin School of Grinds
Page 12
Kieran Mills & Tony Kelly
u
s = 100 m
s = 100 m
a
80 = 6u + 45a.....(2)
200 = 15u +
225
2
200 = u (15) + 12 a (15) 2
t = 15 s, s = 200 m
t = 10 s
t=5s
a
40 = 2u + 5a.....(1)
100 = 5u + 252 a
100 = u (5) + 12 a (5) 2
s = ut + 12 at 2
t=?
In these types of problems always use the equation shown.
Take all values from the beginning point.
0ms
-1
70
3
=u
∴u =
140
6
m s −1
120 = 6u − 20
140 = 6u
40 = 2u − 203
40 = 2u + 5(− 34 )
40 = 2u + 5a.......(1)
− 40 = 30a
−120 = −6u − 15a
80 = 6u + 45a
40 = 2u + 5a.......(1)(× − 3)
80 = 6u + 45a.....(2)
Mathematical Calculations
Example 2
A uniformly decelerating body covers successive 100 m distances in 5 s and 10 s. Find its initial speed, the
deceleration and the further time for the body to come to rest.
Solution
Cont....
©The Dublin School of Grinds
Page 13
Kieran Mills & Tony Kelly
s = 100 m
s = 100 m
t = 15 s, s = 200 m
t = 10 s
t=5s
70
3
m s −1
t=
+ (− 34 )t
70
3
70
3
t=
70
4
= 17.5 s
4t = 70
4
3
0=
v = u + at
Answer: Further time = 17.5 – 15 = 2.5 s
v = 0 m s −1
t =?
a = − 34 m s −2
u=
Further time for body to come to rest:
Mathematical Calculations
u
a
s = ut + 12 at 2
t=?
0 m s-1
70
3
=u
∴u =
140
6
m s −1
120 = 6u − 20
140 = 6u
40 = 2u − 203
40 = 2u + 5(− 34 )
40 = 2u + 5a.......(1)
− 40 = 30a
−120 = −6u − 15a
80 = 6u + 45a
40 = 2u + 5a.......(1)(× − 3)
80 = 6u + 45a.....(2)
Mathematical Calculations
Example 2
A uniformly decelerating body covers successive 100 m distances in 5 s and 10 s. Find its initial speed, the
deceleration and the further time for the body to come to rest.
Solution
Exercise 3. Successive Times/Successive Distances
1. In two successive seconds a uniformly accelerating body travels 4 m and 8 m.
Find its acceleration.
2. A uniformly accelerating body travels 5 m and 11 m repectively in its first two seconds.
How far does it travel in the fourth second?
3. A uniformly decelerating body covers successive 100 m distances in 5 s and 10 s.
Find its initial speed, the deceleration and the further time for the body to come to rest.
4. A particle starts from rest and moves in a straight line with uniform acceleration.
It passes three points A, B and C where |AB| =105 m and |BC| = 63 m. If it takes 6 s to travel
from A to B and 2 s from B to C find
(i) its acceleration, (ii) the distance of A from the starting position.
5. A sprinter runs a race with constant acceleration throughout. During the race he passes four
posts A, B, C, D such that |AB| = |BC| = |CD| = 36 m. If the sprinter takes 3 s to run from A to
B and 2 s to run from B to C, how long does it takes to run from C to D?
6. A particle moving in a straight line with uniform acceleration describes 23 m in the fifth
second of its motion and 31 m in the seventh second. Calculate its initial velocity.
7. A body travels in a straight line with uniform acceleration. The particle passes three points A,
B and C at t = 0, t = 3 s and t = 6 s. If |BC| = 90 m and the speed of the particle at B is
21 m s–1, find the acceleration of the body and its speed at A.
8. A, B, C are three points which lie in that order on a straight road with |AB| = 45 m and
|BC| = 32 m. A car travels along the road in the direction ABC with constant acceleration f.
The car passes A with speed u and passes B five seconds later and passes C two seconds after
that. Find u and f.
9. A car is moving along at a steady 20 m s–1 when the driver suddenly sees a tree across the
road 56 m ahead. He immediately applies the brakes giving the car a constant deceleration of
4 m s–2. How far in front of the tree does the car come to rest? If the driver had not reacted
immediately and the brakes were applied one second later with what speed would the car
have hit the tree?
10. A, B, C are three points on a straight line in that order. A body is projected from B towards A
with a speed of 5 m s–1. The body experiences an acceleration of 2 m s–2 towards C.
If |BC| = 24 m, find the time to reach C, and the distance travelled by the body from the
instant of projection until it reaches C.
11. A bus 12.5 m long travels with constant acceleration. The front of the bus passes a point P
with speed u and the rear passes P with speed v. Find in terms of u and v
(i) the time taken for the bus to pass P,
(ii) what fraction of the bus passes P in half this time.
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
12. A body moving in a straight line with constant acceleration passes in succession through
points A, B, C and D where |AB| = x, |BC| = y and |CD| = z where the distances x, y and z are
covered in equal intervals of time. Show 2y = x + z.
13. A uniformly decelerating train of length 40 m enters a station of length 80 m. The front
engine leaves the station 5 s later and the rear of the train leaves the station after a further
5 s. Find the deceleration of the train.
14. A uniformly accelerating body starts with a speed of u, in successive times of t travels
4u 2
.
distances s and 2s. Prove that its acceleration is
s
15. A body starts moving in a straight line with velocity u and acceleration a. If when the
velocity has increased to 5u the acceleration is reversed in direction its magnitude being
unaltered prove that when the particle returns to its starting point its velocity will be –7u.
Answers
Exercise 3
1. 4 m s–2
2. 23 m
3.
70
3
m s −1 , − 43 m s −2 , 2.5 s
4. (i) 3.5 m s–2, (ii) 7 m
7. 6 m s–2, 3 m s–1
8. 4 m s–1, 2 m s–2
9. 6 m, 10.6 m s–1
10. 8 s, 36.5 m
5. 1.6 s
6. 5 m s
–1
©The Dublin School of Grinds
13. –1.6 m s–2
11.
25
3u + v
,
v + u 4(u + v)
Page 15
Kieran Mills & Tony Kelly
©The Dublin School of Grinds
Page 16
Kieran Mills & Tony Kelly
Two bodies move in the same direction along parallel paths. A starts
from a point O with velocity 8 m s–1 and acceleration 2 m s–2 and B starts
8 m ahead of A and moves off with velocity 2 m s–1 at acceleration
4 m s–2. Find when they will be together and their distances from O at
these times. What are their respective speeds when they are together?
Example 3
Two bodies A and B travel in the same direction along the same line.
Body A starts with velocity 5 m s–1 and acceleration 3 m s–2. The other
body starts from the same place with velocity 2 m s–1 and acceleration
4 m s–2. Find when and where they are together again.
Example 2
Two bodies start together at the same time at the same place and move
along the same straight line. If one moves with a constant speed of
16 m s–1 while the other starts from rest and moves at a constant
acceleration of 4 m s–2. How long will it take before they are together?
Example 1
Section 3: Catch-up Problems
A
64 m
B
u = 1 m s–1
a = 4 m s–2
If A starts 2 seconds before B find when and where they are together.
Find the maximum distance that A moves ahead of B in the subsequent
motion.
u = 10 m s–1
a = 2 m s–2
Example 4
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
B
t
t
sA = sB
s A − sB = 16t − 2t 2
= 2t 2
sB = (0)t + 12 (4)t 2
= 16t
s A = 16t + 12 (0)t 2
Separation Equation: s A − sB
u=0ms
a = 4 m s-2
-1
u = 16 m s-1
-2
A a=0ms
t=0s
s = ut + 12 at 2
Answers
A and B are together after 0 s
and after 8 s.
t 2 − 8t = 0
t (t − 8) = 0
t = 0 s, 8 s
2t 2 − 16t = 0
− sB = 0(level): s A − sB = 0
Ts Aogether
16t − 2t 2 = 0
Example 1
Two bodies start together at the same time at the same place and move along the same straight line. If one moves
with a constant speed of 16 m s–1 while the other starts from rest and moves at a constant acceleration of 4 m s–2.
How long will it take before they are together?
Solution
Mathematical Calculations
©The Dublin School of Grinds
Page 18
Kieran Mills & Tony Kelly
B
t
t
sA = sB
= 3t − 12 t 2
s A − sB = 5t + t − 2t − 2t
3 2
2
2
= 2t + 2t 2
sB = (2)t + 12 (4)t 2
= 5t + 32 t 2
s A = (5)t + 12 (3)t 2
Separation Equation: s A − sB
u = 2 m s-1
a = 4 m s-2
u = 5 m s-1
-2
A a=3ms
t=0s
s = ut + 12 at 2
= 2(6) + 2(6) 2
= 2(6) + 2(36)
= 12 + 72
= 84 m
sB = 2t + 2t 2
Distance A and B have travelled
after 6 s:
A and B are together after 0 s
and after 6 s.
t (t − 6) = 0
t = 0 s, 6 s
t 2 − 6t = 0
3t − 12 t 2 = 0
Together (level): s A − sB = 0
Example 2
Two bodies A and B travel in the same direction along the same line. Body A starts with velocity 5 m s–1 and
acceleration 3 m s–2. The other body starts from the same place with velocity 2 m s–1 and acceleration 4 m s–2. Find
when and where they are together again.
Solution
Mathematical Calculations
©The Dublin School of Grinds
Page 19
Kieran Mills & Tony Kelly
8m
u = 8 m s-1
-2
A a=2ms
t=0s
o
t
= 8 + 2t + 2t 2
sB = 8 + (2)t + 12 (4)t 2
= 8t + t 2
s A = (8)t + 12 (2)t 2
B starts 8 m ahead of A. Therefore, add
8 to the distance equation for B.
= −t 2 + 6t − 8
s A − sB = 8t + t 2 − 8 − 2t − 2t 2
Separation Equation: s A − sB
u = 2 m s-1
a = 4 m s-2
B
t
sA = sB
s = ut + 12 at 2
t 2 − 6t + 8 = 0
(t − 2)(t − 4) = 0
t = 2 s, 4 s
−t 2 + 6t − 8 = 0
Cont.....
Together (level): s A − sB = 0
Mathematical Calculations
Example 3
Two bodies move in the same direction along parallel paths. A starts from a point O with velocity 8 m s–1 and
acceleration 2 m s–2 and B starts 8 m ahead of A and moves off with velocity 2 m s–1 at acceleration 4 m s–2. Find
when they will be together and their distances from O at these times. What are their respective speeds when they
are together?
Solution
©The Dublin School of Grinds
Page 20
Kieran Mills & Tony Kelly
-1
8m
u=8ms
-2
A a=2ms
o
t=0s
u = 2 m s-1
a = 4 m s-2
B
sA = sB = 20 m
= 8 + 2t + 2t 2
vB = 10 m s
-1
sB = 8 + (2)t + 12 (4)t 2
vA = 12 m s
-1
= 8t + t
2
s A = (8)t + 12 (2)t 2
t=2s
t=4s
= 10 m s −1
= (2) + (4)(2)
= 2+8
vB = u + at
= 12 m s −1
= (8) + (2)(2)
=8+ 4
v A = u + at
= 8(2) + (2) 2
= 16 + 4
= 20 m
Cont.....
Example 3
Two bodies move in the same direction along parallel paths. A starts from a point O with velocity 8 m s–1 and
acceleration 2 m s–2 and B starts 8 m ahead of A and moves off with velocity 2 m s–1 at acceleration 4 m s–2. Find
when they will be together and their distances from O at these times. What are their respective speeds when they
are together?
Mathematical Calculations
Solution
Distances and speeds after 2 s:
v = u + at
s = ut + 12 at 2
s A = 8t + t 2
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
-1
8m
u=8ms
-2
A a=2ms
o
t=0s
u = 2 m s-1
a = 4 m s-2
B
sA = sB = 20 m
-1
vB ==10
8m
+ 2st + 2t 2
sB = 8 + (2)t + 12 (4)t 2
vA = 12 m s
-1
= 8t + t 2
s A = (8)t + 12 (2)t 2
t=2s
sA = sB = 48 m
vB = 18 m s-1
vA = 16 m s
-1
t=4s
= 8(4) + (4) 2
= 32 + 16
= 48 m
= 18 m s −1
= (2) + (4)(4)
= 2 + 16
vB = u + at
= 16 m s −1
= (8) + (2)(4)
=8+8
v A = u + at
A
Example 3
Two bodies move in the same direction along parallel paths. A starts from a point O with velocity 8 m s–1 and
acceleration 2 m s–2 and B starts 8 m ahead of A and moves off with velocity 2 m s–1 at acceleration 4 m s–2. Find
when they will be together and their distances from O at these times. What are their respective speeds when they
are together?
Mathematical Calculations
Solution
Distances and speeds after 4 s:
2
1
v = u + at
s = ut + 2 at
s = 8t + t 2
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
-1
64 m
sA = sB
-1
t
t+2
sB = 64 + (1)t + 12 (4)t 2
sB ==264
+ (1)t + 1 (4)t 2
t 2 + t + 64 2
= 2t 2 + t + 64
s A = (10)(t + 2) + 12 (2)(t + 2) 2
s A = (10)(t + 2) + 12 (22)(t + 2) 2
= 10t + 20 + (t + 2)
= 10t + 20 + (t + 2) 2
= 10t + 20 + t 2 + 4t + 4
= 10t + 20 + t 2 + 4t + 4
= t 2 + 14t + 24
= t 2 + 14t + 24
= −t 2 + 13t − 40
s A − sB = t 2 + 14t + 24 − 2t 2 − t − 64
Separation Equation: s A − sB
t=0s
u=1ms
-2
B a=4ms
u = 10 m s
A a = 2 m s-2
t=2s
s = ut + 12 at 2
If A starts 2 s before B, take A’s time as (t + 2) s
and B’s time as t s.
Example 4
If A starts 2 seconds before B find when and where they are together. Find
the maximum distance that A moves ahead of B in the subsequent motion.
Solution
Cont.....
A and B are together (level) 5 s
and 8 s after B starts.
t 2 − 13t + 40 = 0
(t − 5)(t − 8) = 0
t = 5 s, 8 s
−t 2 + 13t − 40 = 0
Together (level): s A − sB = 0
Mathematical Calculations
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
A
64 m
sA = sB = 119 m
t=0s
t
t+2
t=5s
sB = 2t2 + t + 64
2
sA = t + 14t + 24
(t + 2) = 7 s
sA = sB = 200 m
t=8s
(t + 2) = 10 s
Separation Equation: s A − sB = −t 2 + 13t − 40
u = 1 m s-1
-2
B a=4ms
u = 10 m s-1
-2
a=2ms
(t + 2) = 2 s
s = ut + 12 at 2
Example 4
If A starts 2 seconds before B find when and where they are together. Find
the maximum distance that A moves ahead of B in the subsequent motion.
Solution
= (8) 2 + 14(8) + 24
= 64 + 112 + 24
= 200 m
Distance after 8 s:
s A = t 2 + 14t + 24
= (5) 2 + 14(5) + 24
= 25 + 70 + 24
= 119 m
s A = t 2 + 14t + 24
Cont.....
Mathematical Calculations
Distance after 5 s:
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
A
64 m
sA = sB = 119 m
t=0s
t
t+2
t=5s
2
sB = 2t + t + 64
2
sA = t + 14t + 24
(t + 2) = 7 s
sA = sB = 200 m
t=8s
(t + 2) = 10 s
Separation Equation: s A − sB = −t 2 + 13t − 40
u = 1 m s-1
-2
B a=4ms
u = 10 m s-1
-2
a=2ms
(t + 2) = 2 s
s = ut + 12 at 2
The maximum separation occurs when the velocities of
A and B are equal.
Example 4
If A starts 2 seconds before B find when and where they are together. Find
the maximum distance that A moves ahead of B in the subsequent motion.
Solution
= −(6.5) 2 + 13(6.5) − 40
= 2.25 m
s A − sB = −t 2 + 13t − 40
2t + 14 = 4t + 1
13 = 2t
∴ t = 6.5 s
v A = vB
= 1 + (4)t
= 4t + 1
vB = u + at
= 10 + (2)(t + 2)
= 10 + 2t + 4
= 2t + 14
v A = u + at
Mathematical Calculations
Maximum Separation: v A = vB
Exercise 4. Catch up
1. Two bodies start together at the same time at the same place and move along the same
straight line. If one moves with a constant speed of 8 m s–1 while the other starts from rest
and moves at a constant acceleration of 2 m s–2. How long will it take before they are
together?
2. A car A passes a point P on a straight road at a constant speed of 10 m s–1. At the same time
another car B starts from rest at P with uniform acceleration 2.5 m s–2.
(i) When and how far from P will B overtake A.
(ii) If B ceases to accelerate on overtaking, what time elapses between the two cars passing a
point Q which is 3 km from P.
3. A boy runs at 4 m s–1 away from a cyclist who starts at rest and accelerates at
2 m s–2. If the boy has an initial lead of 5 m, how long does the cyclist take to catch him?
4. Two bodies A and B travel in the same direction along the same line. Body A starts with
velocity 3 m s–1 and acceleration 2 m s–2. The other body starts from the same place with
velocity 1 m s–1 and acceleration 3 m s–2. Find when and where they are together again.
5. Two bodies move along parallel tracks in the same direction. Body A starts with velocity
2 m s–1 and acceleration 6 m s–2. Body B starts from the same place and the same time with
velocity 5 m s–1 and acceleration 2 m s–2. Find when and where they are together again. Find
their velocities when they are together for the second time.
6. Two bodies move in the same direction along parallel paths. A starts from point O with
velocity 2 m s–1 and acceleration 4 m s–2. B starts 6 m ahead of A with velocity 3 m s–1 and
acceleration 2 m s–2. Find when and where they are together and their velocities at this
instant.
7. Two bodies move in the same direction along parallel paths. A starts from a point O with
velocity 8 m s–1 and acceleration 2 m s–2 and B starts 8 m ahead of A and moves off with
velocity 2 m s–1 at acceleration 4 m s–2. Find when they will be together and their distances
from O at these times.
8. Two bodies A and B travel in the same direction along the same line from the same point P at
the same time. A starts with velocity 5 m s–1 and acceleration 3 m s–2. B starts with velocity
2 m s–1 and acceleration 4 m s–2. They are together again at point Q. Find the time at which
they are together and the distance | PQ |. Find their maximum distance apart between P and Q.
9. Two bodies move in the same direction along parallel paths. They start at the same point P at
the same time. A starts from P with velocity 3 m s–1 and acceleration 2 m s–2. B starts with
velocity 1 m s–1 and acceleration 3 m s–2. They are together again at point Q. Find the time at
which they are together and the distance | PQ |.Find their maximum distance apart between P
and Q.
10. Two bodies A and B move along parallel straight lines in the same direction from the same
point P. A starts with velocity 4 m s–1 and acceleration 2 m s–2. B starts 1 second after A with
velocity 2 m s–1 and acceleration 4 m s–2. Find when and where they will be together.
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
11. Two bodies A and B move along parallel straight lines in the same direction from the same
point P. A starts from point P with velocity 5 m s–1 and acceleration 4 m s–2. B starts 1
second before A with velocity 6 m s–1 and acceleration 3 m s–2 from a point a distance of
2.5 m to the right of P. Find when and where they are together.
12. Find when and where they are together.
Find their maximum separation between the
two times when they are together.
A
u = 10 m s–1
a = 2 m s–2
u = 1 m s–1
a = 4 m s–2
B
8m
13. If A starts 2 seconds before B find when and
where they are together. Find their maximum
separation between the two positions.
u = 10 m s–1
a = 2 m s–2
A
u = 1 m s–1
–2
B a=4ms
64 m
14. A car A starts from a point P with initial velocity 8 m s–1 and then travels with uniform
acceleration 4 m s–2. Two seconds later a second car B starts from P with an initial velocity
of 30 m s–2 and then moves with a uniform acceleration of 3 m s–2. Show that after passing A,
B will never be ahead by more than 74 m.
15. Bodies A and B start together and move along the same straight line. A starts with a speed of
10 m s–1 and moves with a constant deceleration, while B starts at 5 m s–1 and accelerates at
4 m s–2. Find the deceleration of A if they meet when the velocity of B is twice that of A.
16. The driver of a car travelling at 20 m s–1 sees a second car 120 m in front travelling in the
same direction at a uniform speed of 8 m s–1.
(a) What is the least uniform retardation that must be applied to the faster car to avoid
collision?
(b) If the actual retardation is 1 m s–2 find
(i) the time interval in seconds for the faster car to reach a point 66 m behind the slower
car,
(ii) the shortest distance between the cars.
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
Answers
Exercise 4
1. 8 s
2. (i) 8 s, 80 m
(ii) 146 s
3. 5 s
4. 4 s, 28 m from starting point
5. 1.5 s, 9.75 m, 11 m s–1, 8 m s–1
6. 3 s, 24 m, 14 m s–1, 9 m s–1
7. 2 s, 4 s, 20 m, 48 m
8. 6 s, 84 m, 4.5 m
9. 4 s, 28 m, 2 m
10. 5 s after B starts, 60 m
11. 10 s after A starts, 250 m from P
12. 1 s, 8 s, 12.25 m
13. 5 s and 8 s after B starts, 119 m from A and 200 m from A, 2.25 m
15. 4 m s–2
16. (a) 0.6 m s–2
(b) (i) 6 s, 18 s; (ii) 48 m
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
©The Dublin School of Grinds
Page 28
Kieran Mills & Tony Kelly
A particle P with speed 140 m s–1 begins to decelerate uniformly at a
certain instant while another particle Q starts from rest 6 s later and accelerates uniformly. When the second particle Q has travelled 125 m,
both particles have a speed of 25 m s–1.
(i) Show the motion of both on the same speed-time curve.
(ii) How many seconds after the commencement of deceleration does
the first particle P come to rest?
Example 3
A car travels from A to B. It starts from rest at A and accelerates at
2 m s–2 until it reaches a speed of 30 m s–1. It then travels at this speed
for 600 m and then decelerates at 2.5 m s–2 to come to rest at B. Find
(i)
the total time for the journey,
(ii) the distance from A to B,
(iii) the average speed for the journey.
Example 2
A car starting from rest accelerates uniformly over 8 s to a velocity of
16 m s–1. It then maintains a constant velocity for the next 20 s. It finally
decelerates uniformly to rest for 4 s. Draw a velocity-time curve to represent the motion of the car. Use the graph to find
(i)
the acceleration and distance travelled by the car in the first 8 s,
(ii) the distance travelled by the car over the next 20 s,
(iii) the deceleration and distance travelled by the car over the last 4 s,
(iv) the average velocity of the car for its entire journey.
Example 1
Section 4: Velocity-time Graphs
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
0
2
4
6
8
10
12
14
16
v (m s-1)
4
8
12
16
20
24
28
t (s)
32
Velocity Time Curves
Acceleration = Slope of curve
Distance = Area under curve
Cont....
Example 1
A car starting from rest accelerates uniformly over 8 s to a velocity of
16 m s–1. It then maintains a constant velocity for the next 20 s. It finally decelerates uniformly to rest for 4 s. Draw
a velocity-time curve to represent the motion of the car. Use the graph to find
(i)
the acceleration and distance travelled by the car in the first 8 s,
(ii) the distance travelled by the car over the next 20 s,
(iii) the deceleration and distance travelled by the car over the last 4 s,
(iv) the average velocity of the car for its entire journey.
Solution
Page 30
Run = 8 s
Base b = 8
4
64 m
8
12
16
-1
Rise = 16 m s
Height h = 16
Area of a triangle = 12 bh
0
2
4
6
8
10
12
14
16
v (m s-1)
20
24
28
t (s)
32
Cont....
s = 12 bh = 12 (8)(16) = 64 m
Mathematical Calculations
Rise 16 m s −1
=
= 2 m s −2
(i) a =
Run
8s
Velocity Time Curves
Acceleration a = Slope of curve
Distance s = Area under curve
Example 1
A car starting from rest accelerates uniformly over 8 s to a velocity of
16 m s–1. It then maintains a constant velocity for the next 20 s. It finally decelerates uniformly to rest for 4 s. Draw
a velocity-time curve to represent the motion of the car. Use the graph to find
(i)
the acceleration and distance travelled by the car in the first 8 s,
(ii) the distance travelled by the car over the next 20 s,
(iii) the deceleration and distance travelled by the car over the last 4 s,
(iv) the average velocity of the car for its entire journey.
Solution
m s -2
a=
2
©The Dublin School of Grinds
Kieran Mills & Tony Kelly
0
2
4
6
8
10
12
14
16
v (m s )
-1
4
64 m
8
20
Length l = 20
16
320 m
24
Area of a rectangle = l × b
12
Breadth b = 16
28
t (s)
32
Cont....
(ii) s = l × b = (20)(16) = 320 m
s = 12 bh = 12 (8)(16) = 64 m
Mathematical Calculations
Rise 16 m s −1
=
= 2 m s −2
(i) a =
Run
8s
Velocity Time Curves
Acceleration a = Slope of curve
Distance s = Area under curve
Example 1
A car starting from rest accelerates uniformly over 8 s to a velocity of
16 m s–1. It then maintains a constant velocity for the next 20 s. It finally decelerates uniformly to rest for 4 s. Draw
a velocity-time curve to represent the motion of the car. Use the graph to find
(i)
the acceleration and distance travelled by the car in the first 8 s,
(ii) the distance travelled by the car over the next 20 s,
(iii) the deceleration and distance travelled by the car over the last 4 s,
(iv) the average velocity of the car for its entire journey.
Solution
s -2
2m
Page 31
a=
©The Dublin School of Grinds
Kieran Mills & Tony Kelly
Page 32
4
8
12
16
20
320 m
Run = 4 s
Base b = 4
28
t (s)
32
Area of a triangle = 12 bh
24
32 m
-2
0
64 m
Rise = 16 m s-1
Height h = 16
s
2
4
6
8
10
12
14
16
v (m s-1)
Rise
16 m s −1
=−
= −4 m s −1
Run
4s
Cont....
s = 12 bh = 12 (4)(16) = 32 m
(iii) a =
(ii) s = l × b = (20)(16) = 320 m
s = 12 bh = 12 (8)(16) = 64 m
Mathematical Calculations
Rise 16 m s −1
=
= 2 m s −2
(i) a =
Run
8s
Velocity Time Curves
Acceleration a = Slope of curve
Distance s = Area under curve
Example 1
A car starting from rest accelerates uniformly over 8 s to a velocity of
16 m s–1. It then maintains a constant velocity for the next 20 s. It finally decelerates uniformly to rest for 4 s. Draw
a velocity-time curve to represent the motion of the car. Use the graph to find
(i)
the acceleration and distance travelled by the car in the first 8 s,
(ii) the distance travelled by the car over the next 20 s,
(iii) the deceleration and distance travelled by the car over the last 4 s,
(iv) the average velocity of the car for its entire journey.
Solution
m s -2
a=
2
©The Dublin School of Grinds
4m
a=-
Kieran Mills & Tony Kelly
Page 33
4
8
12
16
20
24
64 m + 320 m + 32 m = 416 m
28
s
t (s)
32
-2
0
2
4
6
8
10
12
14
16
v (m s )
-1
Total Distance
Total Time
(iv) Average velocity =
Cont....
416 m
= 13 m s −1
32 s
Mathematical Calculations
Average Velocity =
Example 1
A car starting from rest accelerates uniformly over 8 s to a velocity of
16 m s–1. It then maintains a constant velocity for the next 20 s. It finally decelerates uniformly to rest for 4 s. Draw
a velocity-time curve to represent the motion of the car. Use the graph to find
(i)
the acceleration and distance travelled by the car in the first 8 s,
(ii) the distance travelled by the car over the next 20 s,
(iii) the deceleration and distance travelled by the car over the last 4 s,
(iv) the average velocity of the car for its entire journey.
Solution
m s -2
a=
2
©The Dublin School of Grinds
4m
a=-
Kieran Mills & Tony Kelly
Page 34
-1
0
8
12
16
20
24
28
t (s)
32
32
= (52)(8)
= 416 m
= 12 (52)(16)
s = 12 (20 + 32)(16)
Area = 12 ( x + y )h
2
4
-2
Mathematical Calculations
16
416 m
20
s
4
6
8
10
12
14
16
v (m s )
Finding the area under the curve in one go:
A trapezium is a four sided shape where two of the
sides are parallel. The area of a trapezium is half the
sum of the parallel sides by the perpendicular distance
between them.
Example 1
A car starting from rest accelerates uniformly over 8 s to a velocity of
16 m s–1. It then maintains a constant velocity for the next 20 s. It finally decelerates uniformly to rest for 4 s. Draw
a velocity-time curve to represent the motion of the car. Use the graph to find
(i)
the acceleration and distance travelled by the car in the first 8 s,
(ii) the distance travelled by the car over the next 20 s,
(iii) the deceleration and distance travelled by the car over the last 4 s,
(iv) the average velocity of the car for its entire journey.
Solution
m s -2
a=
2
©The Dublin School of Grinds
4m
a=-
Kieran Mills & Tony Kelly
©The Dublin School of Grinds
Page 35
Kieran Mills & Tony Kelly
Total time T = 15 s + 20 s + 12 s = 47 s
Example 2
A car travels from A to B. It starts from rest at A and accelerates
at 2 m s–2 until it reaches a speed of 30 m s–1. It then travels at this
speed for 600 m and then decelerates at 2.5 m s–2 to come to rest
at B. Find
(i) the total time for the journey,
(ii) the distance from A to B,
(iii) the average speed for the journey.
Solution
30
= 15 s
2
III : a =
600
= 20 s
30
t3 =
30
= 12 s
2.5
Cont.....
V
30
⇒ 2.5 =
t3
t3
t2 =
s
600
⇒ 30 =
t2
t2
t2 = ?
s s = 600 m
t V = 30 m s −1
II :V =
v=
t1 =
Mathematical Calculations
V
30
(i) I : a = ⇒ 2 =
t1
t1
Velocity Time Curves
Acceleration a = Slope of curve
Distance s = Area under curve
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
Total time T = 15 s + 20 s + 12 s = 47 s
Average Velocity =
Area = 12 ( x + y )h
Total Distance
Total Time
(iii) Average velocity =
1005 m
= 21.4 m s −1
47 s
(ii) S = 12 (20 + 47)30 = 1005 m
Mathematical Calculations
Example 2
A car travels from A to B. It starts from rest at A and accelerates
at 2 m s–2 until it reaches a speed of 30 m s–1. It then travels at this
speed for 600 m and then decelerates at 2.5 m s–2 to come to rest at
B. Find
(i) the total time for the journey,
(ii) the distance from A to B,
(iii) the average speed for the journey.
Solution
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Kieran Mills & Tony Kelly
∴ t = 10 s
5 = 12 t
125 = 12 t (25)
s = 12 bh
h = 25 m s −1
b=t
Mathematical Calculations
(ii) Particle Q: s = 125 m
Example 3
A particle P with speed 140 m s–1 begins to decelerate uniformly at a certain instant while another particle Q starts
from rest 6 s later and accelerates uniformly. When the second particle Q has travelled 125 m, both particles have a
speed of 25 m s–1.
(i) Show the motion of both on the same speed-time curve.
(ii) How many seconds after the commencement of deceleration does the first particle P come to rest?
Solution
Velocity Time Curves
Acceleration a = Slope of curve
Distance s = Area under curve
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
t =?
v = 0 m s −1
a = −7.18 m s −2
−1
v = u + at
0 = 140 + (−7.18)t
7.18t = 140
140
t=
= 19.5 s
7.18
Rise
115
=−
= −7.18 m s −2
Run
16
u = 140 m s
a=
Mathematical Calculations
(ii) Particle P:
Example 3
A particle P with speed 140 m s–1 begins to decelerate uniformly at a certain instant while another particle Q starts
from rest 6 s later and accelerates uniformly. When the second particle Q has travelled 125 m, both particles have a
speed of 25 m s–1.
(i) Show the motion of both on the same speed-time curve.
(ii) How many seconds after the commencement of deceleration does the first particle P come to rest?
Solution
Velocity Time Curves
Acceleration a = Slope of curve
Distance s = Area under curve
Exercise 5. Velocity Time Curves
1.
A car is travelling at 72 km h–1 when the brakes are applied producing a retardation of
4 m s–2. How long does it take to stop?
2.
An electric train starts from a station and reaches a speed of 14 m s–1 in 25 s with uniform
acceleration. Sketch the velocity-time graph, and find how far it has gone by the time it
reaches this speed.
3.
An aircraft can take off when it reaches a speed of 180 km h–1. If it attains this speed in 30 s
with uniform acceleration what distance does it require for taking off?
4.
An express train is travelling at 144 km h–1 when its brakes are applied. If these produce a
retardation of 2 m s–2 how long will it take to stop and what distance will it cover in doing
so?
5.
A train starts from rest and attains a speed of 50 km h–1 in 4 minutes with uniform
acceleration. It runs at that speed for 5 minutes and then slows down uniformly to rest in
2 minutes. Draw the velocity-time graph and find the total distance travelled.
6.
Find from the velocity-time graph shown
(i) the acceleration during the first 4 s,
(ii) the retardation during the last 2 s,
(iii) the total distance travelled.
v (m/s)
8
0
4
8 10
t (s)
7.
A cyclist rides along a straight road from A to B. He starts from rest at A and accelerates
uniformly to reach a speed of 10 m s–1 in 8 s. He maintains this speed for 30 s and then
uniformly decelerates to rest at B. If the total time is 48 s, draw a velocity-time curve and
from it find
(i) the acceleration,
(ii) the deceleration,
(iii) the total distance travelled.
8.
A car travels from A to B. It starts from rest at A and accelerates at 1.5 m s–2 until it reaches a
speed of 30 m s–1. It then travels at this speed for 2 km and then decelerates at 2 m s–2 to
come to rest at B. Find
(i) the total time for the journey,
(ii) the distance from A to B,
(iii) the average speed for the journey.
9.
A and B are two points on a straight road. A car travelling along the road passes A when
t = 0 and maintains a constant speed until t = 20 s, and in this time covers four-fifths of the
distance from A to B. The car then decelerates uniformly to rest at B. Draw a velocity-time
curve and find the time from A to B.
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
10. A tram travels along a straight track and starts from rest. It accelerates uniformly for 20 s
and during this time it travels 160 m. It maintains a constant speed for a further 50 s and
decelerates to rest in 8 s.
Calculate
(i) the acceleration,
(ii) the deceleration,
(iii) the total time,
(iv) the total distance.
11. A train starts from rest and travels 8 km in 12 minutes ending at rest. The acceleration is half
the retardation, both are uniform, and there is a period when the train runs at its maximum
speed of 50 km h–1. Find the time taken to reach full speed.
12. A 100 m sprinter starts with a speed of 6 m s-1 and accelerates uniformly to 10 m s–1 and
finishes the race at this speed. If his total time is 10.4 s, find his uniform acceleration and
after what distance he is going at full speed.
13. A car takes 2 minutes to travel between two sets of traffic lights 2145 m apart. It has uniform
acceleration for 30 s, then uniform velocity, and then uniform retardation for the last 15 s.
Find the maximum velocity and the acceleration.
14. A train travels 15 km between two stations at an average speed of 50 km h–1. Its acceleration
is half the retardation and both are uniform. If the maximum speed is 72 km h–1 find the
acceleration in m s–2. Sketch the velocity-time curve.
15. A car accelerates at 2 m s–2 in bottom gear, 1.5 m s–2 in second gear and 1 m s–2 in top gear.
Each gear change takes 1.5 s during which time the car travels at constant speed. If a
motorist changes gear when his speeds are 3 m s–1 and 9 m s–1 find how long he will take to
reach 15 m s–1 from rest.
16. A train moving in a straight line starts from A with uniform acceleration of 0.1 m s–2. After it
has attained full speed it moves uniformly for 10 minutes. It is brought to rest at B by the
brakes, which apply a constant retardation of 0.8 m s–2 for 20 s. Draw a rough velocity-time
graph and from it find the time of the journey and the distance from A to B.
17. A train has a maximum speed of 72 km h–1 which it can achieve at an acceleration of
0.25 m s–2. With its brakes fully applied the train has a deceleration of 0.5 m s–2. What is the
shortest time that the train can travel between stations 8 km apart if it stops at both stations?
18. A particle with speed 150 m s–1 begins to decelerate uniformly at a certain instant while
another particle starts from rest 8 s later and accelerates uniformly. When the second particle
has travelled 135 m, both particles have a speed of 30 m s–1.
(i) Show the motion of both on the same speed-time curve.
(ii) How many seconds after the commencement of deceleration does the first particle come
to rest?
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
19. A body starts from rest at P travelling in a straight line and then comes to rest at Q which is
696 m from P. The time taken is 66 s. For the first 10 s it has uniform acceleration a. It then
travels at constant speed and is finally brought to rest by a uniform deceleration b acting for
6 s. Find a and b.
If the journey from rest at P to rest at Q had been travelled with no interval of constant speed
but at acceleration of a for a time t1 immediately followed by deceleration b for a time t2,
show that the time for the journey is 8 29 s.
20. An athlete runs 100 m in 12 s. Starting from rest he accelerates uniformly to a speed of
10 m s–1 and then continues at that speed. Calculate the acceleration.
21. A cyclist has a maximum acceleration of 2 m s–2, a maximum speed of 15 m s–1 and a
maximum deceleration of 4 m s–2. If he travels from rest to rest in the shortest possible time
show that he covers a distance of 84 83 m. Find the time to travel
(i) 105 m, (ii) 54 m.
Answers
Exercise 5
1. 5 s
14.
2. 175 m
16. 13 minutes, 11.04 km
4. 20 s, 400 m
17. 460 s
5. 6 23 km
6. (i) 2 m s–2
18. (ii) 21.25 s
(ii) 4 m s–2 (iii) 56 m
7. (i) 1.25 m s–2 (ii) 1 m s–2 (iii) 390 m
305
3
m s −2
15. 14.5 s
3. 0.75 km
8. (i)
1
22
s
(ii) 2525 m (iii) 24.83 m s–1
19. 1.2 m s–2, 2 m s–2
20. 2.5 m s–2
21. (i) 12 85 s
(ii) 9 s
9. 30 s
10. (i) 0.8 m s–2
(iii) 78 s
(ii) 2 m s–2
(iv) 1024 m
11. 3.2 minutes
12. 2 m s–2, 16 m
13. 22 m s–1, 0.73 m s–2
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
A hot-air balloon travels from the ground vertically up at a constant
speed of 12 m s-1. Find its height above the ground after 5 s.
After 5 s a ball is dropped from the baloon. How long does it take to
reach the ground?
Example 3
A ball is thrown vertically up at 10 m s-1 from a point 3 m above the
ground. Find the speed it has when it hits the ground and the time it
takes the ball to hit the ground.
Example 2
A body is thrown vertically up from the ground at 14 m s-1.
Find the maximum height it reaches and the time to reach this height.
From its highest point find the time for the body to hit the ground and its
speed when it hits the ground.
Example 1
Section 5: Free Fall
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
s=?
v = u + at
0 = 14 − 9.8t
9.8t = 14
14 10
t=
=
= 1.43 s
9.8 7
02 = 142 + 2(−9.8) s
19.6 s = 196
196
∴s =
= 10 m
19.6
v 2 = u 2 + 2as
v=?
v = u + at
−14 = 0 − 9.8t
14 10
t=
= = 1.43 s
9.8 7
∴ v = 196 = −14 m s −1
v 2 = 196
v 2 = 02 + 2(−9.8)(−10)
v 2 = u 2 + 2as
Up is positive
Down is negative
Velocity going up = Velocity at same point on way down but in the opposite direction
s = 10 m
Going down:
u = 0 m s-1
a = -9.8 m s-1
s = -10 m
t=?
v=?
u = 0 m s1
GOING DOWN
The acceleration due to gravity is denoted by g.
This value is 9.8 m s-2.
In free fall alway let a = -g = -9.8 m s-2.
Conclusions: Time up = Time Down
u = 14 m s1
Going up:
u = 14 m s-1
v = 0 m s-1
a = -9.8 m s-1
s=?
t=?
v = 0 m s1
GOING UP
Up is positive
Down is negative
Example 1
A body is thrown vertically up from the ground at 14 m s-1.
Find the maximum height it reaches and the time to reach this height.
From its highest point find its speed when it hits the ground and the time for the body to hit the ground.
Solution
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
v=?
Up is positive
Down is negative
The acceleration due to gravity is denoted by g.
This value is 9.8 m s-2.
In free fall alway let a = -g = -9.8 m s-2.
s = 3 m
u = 10 m s1
u = 10 m s-1
s = -3 m
a = -9.8 m s-1
v=?
t=?
v = u + at
−12.6 = 10 − 9.8t
−22.6 = −9.8t
22.6
t=
= 2.31 s
9.8
The velocity is negative as it is moving down.
v = 102 + 2(−9.8)(−3) = −12.6 m s −1
v 2 = 102 + 2(−9.8)(−3)
v 2 = u 2 + 2as
Example 2
A ball is thrown vertically up at 10 m s-1 from a point 3 m above the ground. Find the speed it has when it hits the
ground and the time it takes the ball to hit the ground.
Solution
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
s=?
v = 12 m s1
t=5s
v = 12 m s1
t=?
u = 12 m s1
Ball is dropped
s = 60 m
v = 12 m s −1
t =5s
s
v = ⇒ s = v × t = 12 × 5 = 60 m
t
12 ± (−12) 2 − 4(4.9)(−60)
= 4.9 s
∴t =
2(4.9)
4.9t 2 − 12t − 60 = 0
−60 = 12t − 4.9t 2
−60 = 12t + 12 (−9.8)t 2
s = ut + 12 at 2
Ball is dropped from balloon:
u = 12 m s-1
s = -60 m
a = -9.8 m s-2
t=?
s
v=
t
Example 3
A hot-air balloon travels from the ground vertically up at a constant speed of 12 m s-1. Find its height above the
ground after 5 s.
After 5 s a ball is dropped from the balloon. How long does it take to reach the ground? Give your answer to one
place of decimal.
Solution
Exercise 6. Uniformly g accelerated motion
[In all problems g = 9.8 m s–2]
1. A vase falls from a shelf 140 cm above the floor. Find the speed with which it strikes the
floor.
2. A stone is dropped from a point 49 m above the ground. Find the time for it to reach the
ground.
3. A stone is thrown down at 5 m s–1. If its speed on hitting the ground is 19 m s–1 from what
height was it thrown. How long does it take?
4. A stone is dropped from the top of a tower and falls to the ground. If it strikes the ground at
14 m s–1, how high is the tower?
5. A ball is thrown vertically downwards from the top of a tower with an initial speed of
2 m s–1. If it hits the ground 3 s later find
(i) the height of the tower,
(ii) the speed with which it hits the ground.
6. A stone is thrown upwards with a speed of 21 m s–1. Find its height
(i) 1 s after projection,
(ii) 2 s after projection,
(iii) 3 s after projection.
7. A ball is thrown up at 14 m s–1 from a point 2 m above the ground. Find
(i) the speed when it returns to the level of projection,
(ii) the speed on the ground.
8. A ball is thrown vertically up at 28 m s–1. Find
(i) the maximum height,
(ii) the time to reach the maximum height,
(iii) the velocity of return,
(iv) the total time for the journey.
9. A balloon is rising at a steady speed of 3 m s–1. How high is it above the ground after 10 s?
At this instant a man releases a stone. What is the initial velocity of the stone?
How long does it take to reach the ground? How high is the balloon above the ground when
the stone strikes the ground?
10. A stone is thrown up at 49 m s–1 from the ground. Find the times at which the particle is
78.4 m above the ground. Find the time interval for which the particle is above 78.4 m.
11. A ball is thrown up at 14 m s–1. Find the times at which the particle is 9.1 m above the
ground.
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
12. A ball is thrown up at 49 m s–1. How long does it take to reach its maximum height? If an
other ball was thrown up 1 s after the first one, how high is it above the ground when the first
ball has reached its maximum height if it has the same initial velocity?
13. A jumper can jump 2 m on the Earth. What is his take-off speed? How high can he jump on
the moon? (Acceleration due to gravity of moon g = 1.6 m s–1 )
14. A particle is thrown vertically upwards under gravity with a speed of 16 m s–1. One second
later another particle is fired upwards from the same point. Find the initial speed of this
particle in order that the two particles will collide when the first particle has reached its
highest point.
15. An object falls vertically past a window 2 m high in
from which the object was dropped.
1
12
s. Find the height above the window
16. A stone is dropped from a balloon rising at 10 m s–1 and reaches the ground in
8 s. How high was the balloon above the ground when the stone was dropped?
17. A body falls from the top of a tower and during the last second it falls
distance. Find the height of the tower.
9
25
of the total
18. A particle falls freely from rest from a point O passing three points A, B and C, the distances
|AB| and |BC| being equal. If the particle takes 3 s to pass from A to B and 2 s from B to C,
calculate |AB|.
19. A body falls freely from rest from a point O passing three points A, B and C, the distances
|AB| and |BC| being equal. The time taken to go from A to B is 2 s and from B to C is 1 s.
Find |AB|.
20. A particle falls freely under gravity from rest at a point P. After it has fallen for 1 s another
particle is projected vertically downwards from P with speed 14.7 m s–1. Find the time and
distance from P at which they collide.
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
Answers
Exercise 6
1. 5.24 m s–1
13. 6.26 m s–1, 12.25 m
2. 3.16 s
14. 23.7 m s–1
3. 17.14 m, 1.43 s
15. 28.4 m
4. 10 m
16. 233.6 m
5.
(i) 50.1 m
6. (i) 16.1 m
(ii) 22.4 m
7. (i) 14 m s–1
(ii) 15.34 m s–1
8. (i) 40 m
(ii)
(iv)
40
7
17. 122.5 m
(ii) 31.4 m s–1
20
7
s
(iii) 18.9 m
18. 147 m
19. 29.4 m
(iii) 28 m s–1
20. 2 s, 19.6 m
s
9. 30 m, 3 m s–1, 2.8 s, 38.4 m
10. 2 s, 8 s, 6 s
11. 1 s,
13
7
s
12. 5 s, 117.6 m
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
Leaving Cert Questions
2014
1. (a) Two cars, P and Q, travel with the same constant velocity 15 m s–1 along a straight level
road. The front of car P is 24 m behind the rear of car Q. At a given instant both cars
decelerate, P at 4 m s–2 and Q at 5 m s–2.
(i)
Find, in terms of t, the distance between the cars t seconds later.
(ii)
Find the distance between the cars when they are at rest.
2013
1. (a) A ball is thrown vertically upwards with a speed of 44·1 m s−1.
Calculate the time interval between the instants that the ball is 39·2 m above the point
of projection.
(b) A lift ascends from rest with constant acceleration f until it reaches a speed v. It
continues at this speed for t1 seconds and then decelerates uniformly to rest with
deceleration f.
The total distance ascended is d, and the total time taken is t seconds.
(i) Draw a speed-time graph for the motion of the lift.
(ii) Show that v =
1
2
f (t - t1 ).
(iii) Show that t1 = t 2 −
©The Dublin School of Grinds
4d
.
f
Page 49
Kieran Mills & Tony Kelly
2012
1. (a) A particle falls from rest from a point P. When it has fallen a distance 19·6 m a second
particle is projected vertically downwards from P with initial velocity 39·2 m s–1.
The particles collide at a distance d from P.
Find the value of d.
(b) A car, starts from rest at A, and accelerates uniformly at 1 m s–2 along a straight level road
towards B, where AB = 1914 m. When the car reaches its maximum speed of 32 m s–1, it
continues at this speed for the rest of the journey.
At the same time as the car starts from A a bus passes B travelling towards A with a
constant speed of 36 m s–1. Twelve seconds later the bus starts to decelerate uniformly at
0·75 m s–2.
(i) The car and the bus meet after t seconds. Find the value of t.
(ii) Find the distance between the car and the bus after 48 seconds.
2011
1. (a) A particle is released from rest at A and falls vertically passing
two points B and C.
2t
It reaches B after t seconds and takes
seconds to fall from
7
B to C, a distance of 2.45 m.
A
B
Find the value of t.
(b) A car accelerates uniformly from rest to a speed v in t1 seconds.
It continues at this constant speed for t seconds and then decelerates
uniformly to rest in t2 seconds.
The average speed for the journey is
C
3v
.
4
(i) Draw a speed-time graph for the motion of the car.
(ii) Find t1 + t2 in terms of t.
(iii) If a speed limit of
2v
were to be applied, find in terms of t the least time the
3
journey would have taken, assuming the same acceleration and deceleration as in
part (ii).
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
2010
1. (a) A car is travelling at a uniform speed of 14 ms–1 when the driver notices a traffic light
turning red 98 m ahead.
Find the minimum constant deceleration required to stop the car at the traffic light,
(i) if the driver immediately applies the brake
(ii) if the driver hesitates for 1 second before applying the brake.
(b) A particle passes P with speed 20 ms–1 and moves in a straight line to Q with uniform
acceleration.
In the first second of its motion after passing P it travels 25 m.
In the last 3 seconds of its motion before reaching Q it travels 13
20 of |PQ|.
Find the distance from P to Q.
2009
1. (a) A particle is projected vertically upwards from
the point P. At the same instant a second particle
is let fall vertically from Q.
The particles meet at R after 2 seconds.
Q
R
The particles have equal speeds when they meet
at R.
P
Prove that |PR| = 3|RQ|.
(b) A train accelerates uniformly from rest to a speed v m/s with uniform
acceleration f m/s2.
It then decelerates uniformly to rest with uniform retardation 2f m/s2.
The total distance travelled is d metres.
(i) Draw a speed-time graph for the motion of the train.
(ii) If the average speed of the train for the whole journey is
©The Dublin School of Grinds
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d
, find the value of f.
3
Kieran Mills & Tony Kelly
2008
1. (a) A ball is thrown vertically upwards with an initial velocity of 39.2 m/s. Find
(i) the time taken to reach the maximum height
(ii) the distance travelled in 5 seconds.
(b) Two particles P and Q, each having constant acceleration, are moving in the same
direction along parallel lines. When P passes Q the speeds are 23 m/s and 5.5 m/s,
respectively. Two minutes later Q passes P, and Q is then moving at 65.5 m/s. Find
(i) the acceleration of P and the acceleration of Q
(ii) the speed of P when Q overtakes it
(iii) the distance P is ahead of Q when they are moving with equal speeds.
2007
1. (a) A particle is projected vertically downwards from the top of a
tower with speed u m/s.
It takes the particle 4 seconds to reach the bottom of the tower.
During the third second of its motion the particle travels 29.9 metres.
t=0
t=2
29.9 m
Find
(i) the value of u
t=3
(ii) the height of the tower.
(b) A train accelerates uniformly from rest to a speed v m/s.
It continues at this speed for a period of time and then decelerates uniformly to rest.
In travelling a total distance d metres the train accelerates through a distance
pd metres and decelerates through a distance qd metres, where p < 1 and q < 1.
(i) Draw a speed-time graph for the motion of the train.
(ii) If the average speed of the train for the whole journey is
find the value of b.
©The Dublin School of Grinds
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v
,
p+q+b
Kieran Mills & Tony Kelly
2006
1. (a) A lift starts from rest. For the first part of its descent it travels with uniform
acceleration f. It then travels with uniform retardation 3f and comes to rest.
The total distance travelled is d and the total time taken is t.
(i) Draw a speed-time graph for the motion.
(ii) Find d in terms of f and t.
(b) Two trains P and Q, each of length 79.5 m, moving in opposite directions along parallel
lines, meet at O, when their speeds are 15 m/s and 10 m/s respectively.
The acceleration of P is 0.3 m/s2 and the acceleration of Q is 0.2 m/s2.
It takes the trains t seconds to pass each other.
(i) Find the distance travelled by each train in t seconds.
(ii) Hence, or otherwise, calculate the value of t.
(iii) How long does it take for
2
5
of the length of train Q to pass the point O?
2005
1. (a) Car A and car B travel in the same direction along a horizontal straight road.
Each car is travelling at a uniform speed of 20 m/s.
Car A is at a distance of d metres in front of car B.
At a certain instant car A starts to brake with a constant retardation of 6 m/s2.
0.5 s later car B starts to brake with a constant retardation of 3 m/s2 .
Find
(i) the distance travelled by car A before it comes to rest
(ii) the minimum value of d for car B not to collide with car A.
2004
1. (a) A ball is thrown vertically upwards with an initial velocity of 20 m/s.
One second later, another ball is thrown vertically upwards from the same point with an
initial velocity of u m/s.
The balls collide after a further 2 seconds.
(i) Show that u = 17.75.
(ii) Find the distance travelled by each ball before the collision, giving your answers
correct to the nearest metre.
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
2003
1. (a) The points P, Q and R all lie in a straight line.
A train passes point P with speed u m/s. The train is travelling with uniform retardation
f m/s2. The train takes 10 seconds to travel from P to Q and 15 seconds to travel from Q
to R, where | PQ | = | QR | = 125 metres.
(i) Show that f = 13 .
(ii) The train comes to rest s metres after passing R.
Find s, giving your answer correct to the nearest metre.
(b) A man runs at constant speed to catch a bus.
At the instant the man is 40 metres from the bus, it begins to accelerate uniformly from
rest away from him.
The man just catches the bus 20 seconds later.
(i) Find the constant speed of the man.
(ii) If the constant speed of the man had instead been 3 m/s, show that the closest he gets
to the bus is 17.5 metres.
2002
1. (a) A stone is thrown vertically upwards under gravity with a speed of u m/s from a point
30 metres above the horizontal ground.
The stone hits the ground 5 seconds later.
(i) Find the value of u.
(ii) Find the speed with which the stone hits the ground.
(b) A particle, with initial speed u, moves in a straight line with constant acceleration.
During the time interval from 0 to t, the particle travels a distance p.
During the time interval from t to 2t, the particle travels a distance q.
During the time interval from 2t to 3t, the particle travels a distance r.
(i) Show that 2q = p + r.
(ii) Show that the particle travels a further distance 2r – q in the time interval from 3t
to 4t.
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
2001
1. (a) Points P and Q lie in a straight line, where | PQ | = 1200 metres.
Starting from rest at P, a train accelerates at 1 m/s2 until it reaches the speed limit of
20 m/s. It continues at this speed of 20 m/s and then decelerates at 2 m/s2, coming to rest
at Q.
Find the time it takes the train to go from P to Q.
Find the shortest time it takes the train to go from rest at P to rest at Q if there is no
speed limit, assuming that the acceleration and deceleration remain unchanged at 1 m/s2
and 2 m/s2, respectively.
(b) A particle is projected vertically upwards with an initial velocity of u m/s and
another particle is projected vertically upwards from the same point and with the same
initial velocity T seconds later.
Show that the particles
T u 
(i) will meet  +  seconds from the instant of projection of the first particle
2 g
4u 2 − g 2T 2
(ii) will meet at a height of
metres.
8g
2000
1. (a) A stone projected vertically upwards with an initial speed of u m/s rises 70 m in the first
t seconds and another 50 m in the next t seconds.
Find the value of u.
(b) A car, starting from rest and travelling from P to Q on a straight level road, where
| PQ | = 10 000 m, reaches its maximum speed 25 m/s by constant acceleration in the first
500 m and continues at this maximum speed for the rest of the journey.
A second car, starting from rest and travelling from Q to P, reaches the same
maximum speed by constant acceleration in the first 250 m and continues at this
maximum speed for the rest of the journey.
(i) If the two cars start at the same time, after how many seconds do the two cars meet?
Find, also, the distance travelled by each car in that time.
(ii) If the start of one car is delayed so that they meet each other exactly halfway
between P and Q, find which car is delayed and by how many seconds.
1999
1 (b) A particle travels in a straight line with constant acceleration f for 2t seconds and covers
15 metres. The particle then travels a further 55 metres at constant speed in 5t seconds.
Finally the particle is brought to rest by a constant retardation 3f.
(i) Draw a speed-time graph for the motion of the particle.
(ii) Find the initial velocity of the particle in terms of t.
(iii) Find the total distance travelled in metres, correct to two decimal places.
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
1998
1 (a) A train accelerates uniformly from rest to a speed v m/s. It continues at this constant
speed for a period of time and then decelerates uniformly to rest. If the average speed for
the whole journey is 56 v, find what fraction of the whole distance is described at constant
speed.
(b) Car A, moving with uniform acceleration 203 b m/s2 passes a point P with speed 9u m/s.
Three seconds later car B, moving with uniform acceleration 92 b m/s2 passes the same
point with speed 5u m/s. B overtakes A when their speeds are 6.5 m/s and 5.4 m/s
respectively. Find
(i) the value u and the value b,
(ii) the distance travelled from P until overtaking occurs.
1997
1 (a) A particle, moving in a straight line, accelerates uniformly from rest to a speed v m/s. It
continues at this constant speed for a time and then decelerates uniformly to rest, the
magnitude of the deceleration being twice that of the acceleration. The distance travelled
while accelerating is 6 m. The total distance travelled is 30 m and the total time taken
is 6 s.
(i) Draw a speed-time graph and hence, or otherwise, find the value of v.
(ii) Calculate the distance travelled at v m/s.
(b) Two points P and Q are a distance d apart. A particle starts from P and move towards Q
with initial velocity 2u and uniform acceleration f. A second particle starts at the same
time from Q and moves towards P with initial velocity 3u and uniform deceleration f.
Prove that
d
(i) the particles collide after
seconds,
5u
(ii) if the particles collide before the second particle comes to instantaneous rest, then
fd < 15u 2,
(iii) if fd = 30u 2 then the second particle has returned to Q before the collision.
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly
1996
1 (a) A particle starts from rest and moves in a straight line with uniform acceleration. It
passes three points A, B and C where |AB| = 105 m and |BC| = 63 m. If it takes 6 seconds
to travel from A to B and 2 seconds to travel from B to C find
(i) its acceleration
(ii) the distance of A from the starting position.
(b) A lift starts from rest with constant acceleration 4 m/s2. It then travels with uniform
speed and finally comes to rest with constant retardation 4 m/s2. The total distance
travelled is d and the total time taken is t.
(i) Draw a speed-time graph.
(ii) Show that the time for which it travelled with uniform speed is
©The Dublin School of Grinds
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t2 − d .
Kieran Mills & Tony Kelly
Answers
Leaving Cert. Questions
2014.
1
(a) (i) 24 - 12 t 2 , (ii) 18.375 m
2013.
1
(a) 7 s
2012.
1
(a) 44.1 m
(b) (i) 40 s, (ii) 352 m
2011.
1
(a) t = 78 s
(b) (ii) t = t1 + t2, (iii)
2010.
1
(a) (i) –1 m s–2, (ii) − 76 m s −2
(b) 300 m
2009.
1
(b) (ii) f = 1 m s–2
2008.
1
2007.
1 (a) (i) 5.4 m/s, (ii) 100 m; (b) (ii) b = 1
2006.
1 (a) (ii)
31
12
t
(a) (i) 4 s, (ii) 83.3 m
(b) (i) aP 245=
=
m/s 2 , aQ 0.5 m/s 2 (ii) 48 m/s
3
8
(iii) 525 m
ft 2 ;
(b) (i) sP = 15t + 0.15t 2 , sQ = 10t + 0.1t 2, (ii) 6 s, (iii) 3.1 s
2005.
1
(a) (i)
2004.
1
(a) (ii) 25 m, 16 m
2003.
1
(a) (ii) 51 m; (b) (i) 4 m/s
2002.
1
(a) (i) 18.5 m/s, (ii) 30.5 m/s;
2001.
1
(a) 75 s, 60 s;
2000.
1
(a) 56 m/s;
(b) (i) 215 s, P: 4875 m, Q: 5125 m, (ii) Delay car starting at Q by 5 s
1999.
1
(b) (ii)
1998.
1
(a)
100
3
m, (ii) 43.3 m
4
, (iii) 75.76 m
t
4
5
(b) (i) u = 0.1 m/s, b = 1, (ii) 94.5 m
1997.
1 (a) (i) 6.5 m/s, (ii) 21 m;
1996.
1
(a) (i) 3.5 m/s2, (ii) 7 m
©The Dublin School of Grinds
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Kieran Mills & Tony Kelly