ASSIGNMENT SOLUTIONS
1.
1 ms-2
20 m/s
x
96 m
Suppose at x distance from the car, one overtakes the other.
For the car, s = x , u = 0 , a = 1 ms-2
 s = ut + ½ a t2 becomes ;
x = ½ x 2 x t2
; x = t2 ---------- (1)
For the cyclist, s = 96 + x , u = 20 m/s , a = 0
 96 + x = 20 t ----------------(2)
From (1) and (2),
t2 – 20t + 96 = 0
(t – 8) (t -12 ) = 0
i.e. t = 8 s, 12 s
2.
2x
x
x
t1
t2
Let the total distance traveled by the body is 2x.
Time taken to complete half the distance t1 = ( x / 40 ) hr
Time taken to complete the remaining half; t2 = ( x / 60 ) hr
Total distance traveled
Average speed =
2x
=
Total time taken
t 1 + t2
2x
=
= 48 km/h
( x / 40 ) + (x / 60 )
3.
Method same as above.
4.
Distance traveled by a body in the tth second = distance traveled in t seconds –
distance traveled in (t-1) seconds.
Distance travelled by a body in t sec. St = ut + ½ a t2
Distance travelled by a body in t – 1 sec; St-1 = u(t-1) + ½ a (t - 1)2
 Stth = St - St-1
= ut + ½ a t2 - {u(t-1) + ½ a (t - 1)2 }
= ut + ½ a t2 - { ut – u + ½ a t2 – at + ½ a }
= ut + ½ a t2 - ut + u - ½ a t2 + at - ½ a
Stth = u + ½ a (2t – 1 )
5. Let the distance covered be x given by :
x  t2
 x = k t2
velocity v = dx/dt = 2kt
acceleration a = dv/dt = 2k
= constant
 the body is under constant acceleration.
6.
Given that the displacement x = 2t2 + 5t m
velocity v = dx/dt = 4t + 5 m/s
acceleration a = dv/dt = 4 = constant
 the body is in constant acceleration.
7.
500 km/h
1500 km/h
Observer at rest
w.r.t ground
Given : velocity of jet w.r.t the ground Vjg = 500 km/h
velocity of products w.r.t the jet , Vpj = -1500 km/h
To find : velocity of the products w.r.t the observer (ground) ,i.e Vpg
Vpg =
=
=
=
=
Vp - Vg
Vp - Vj + Vj - Vg
Vpj + Vjg
-1500 + 500
- 1000 km/h
8.
v
(m/s)
50
Note : graph not to scale
5
7
10
Distance = Area inside the curve = 300 m
t (s)
9. Method 1 (Recommended) – Using the concept of relative velocity.
20m/s
B
A
20m/s
x
Given :
VAG = 72km/h = 20 m/s
VBG = 72km/h = 20 m/s
VBA = 0 m/s
t = 50 s
Let x be the distance between the driver of A and guard of B.
Let us consider A at rest.
 distance travelled by the guard to ‘see’ the driver is x.
Using the equation s = ut + ½ a t2 , we get, x = 0 + ½ a t2
(initial relative velocity of B w.r.t A = 0 )
 x = ½ x 1 x 502
= 1250 m
Method 2 : ( without using relative velocity )
Let at some distance x’ guard of B ‘says goodbye’ to the driver of A.
Distance travelled by the driver, x’ = ut + ½ a t2
x’ = 20 x 50 ( a = 0 for the train A )
= 1000 m
To say goodbye, the guard must travel a distance = x + x’ in 50 s
 x + x’ = ut + ½ a t2
= 20x50 + ½ x 1 x 502
= 1000 + 1250 = 2250
 the initial distance between the driver and the guard x = 2250 – 1000
= 1250 m
10.
B
A
C
At the moment when B decides to overtake A,
VA = 36 km/h = 10 m/s
VB = 54 km/h = 15 m/s
VC = - 54 km/h = - 15 m/s
Let us think the car A is at rest.
VCA = VC – VA = - 25 m/s
; VBA = 5 m/s
Time taken by C to cross A , t = 1000 / 25 = 40 s
 Car B should overtake A in a time less than 40 s
i.e. using the equation , s = ut + ½ a t2 , we get ,
1000 = 5 x 40 + ½ a (402)
800 = 800 a
;
a = 1 ms-2
11.
8.3m/s
53.3m/s
150m/s
Given :
VV = VVG = 30 km/h = 8.3 m/s
VC = VCG = 192 km/h = 53.3 m/s
VBV = 150 m/s
VCV = VC – VV = 53.3 – 8.3 = 45 m/s
Velocity by which the bullet strikes the car = VBC
VBC = VB – VC
= VB – VV + VV – VC
= VBV + VVC
= 150 + (-45)
= 105 m/s
12.
v (m/s)
Note : Graph not to scale
42
37.8
4.3
8.2
12.1
t (s)
13.
Given :
u = 9.8 m/s (the direction of initial velocity is upwards)
s = - 39.2 m , a = - 9.8 m/s2 ( displacement and acceleration downwards)
Using the equation,
s = ut + ½ a t2
-39.2 = 9.8 t + ½ (-9.8) t2
4.9 t2 – 9.8 t – 39.2 = 0
Solving the quadratic, we get, t = -2 , t = 4
 t = 4 s (since time cant be negative)
Using the equation, v = u + at
= 9.8 – 9.8 x 4
= - 29.4 m/s (neg. sign shows the direction of velocity)
14.
Given :
v = 2t2 + 5 cm/s
velocity at 2 s , v = 2 (22) + 5 = 13 cm/s
velocity at 4 s , v = 2 (42) + 5 = 37 cm/s
 change in velocity = 37 – 13 = 24 cm/s
average acc. = v / t = (37 – 13) / (4 – 2)
= 24 / 2 = 12 cm/s2
acceleration a = dx / dt = 4t
acceleration at 4 s , a = 4 x 4 = 16 cm/s2
15.
Distance travelled by a body in nth second
Sn = u + ½ a (2n – 1)
Using the above equation, we get,
12 = u + ½ a (4 – 1)
12 = u + 3/2 a --------------------- (1)
20 = u + ½ a (8 – 1 )
20 = u + 7/2 a --------------------- (2)
(2) – (1), we get ,
8=2a
 a = 4 m/s2
Substituting a in (1) we get,
u = 12 – 3/2 a
= 12 – 6
= 6 m/s
Distance travelled in 4 sec. after the 5th second = S9 – S5
S9 = ut + ½ a t2
= 6x9 + ½ x 4 x 92
= 54 + 162 = 216 m
S5 = ut + ½ a t2
= 6x5 + ½ x 4 x 52
= 30 + 50 = 80 m
Distance travelled in 4 sec. after the 5th second = S9 – S5
= 216 – 80 = 136 m
16. See the class notes, already solved.
17.
B
20m/s
40m
20m/s
A
Let at a distance x from the ground they collide.
For the body A :
x = 20 t + ½ (-9.8) t2
x = 20t – 4.9 t2 -----------------(1)
For the body B :
-( 40 – x ) = - 20 t + ½ (-9.8) t2
-40 + x = - 20t – 4.9 t2 -----------------(2)
Taking (1) – (2) , we get,
40 = 40 t
 t = 1s
Using (1), the distance x = 20 – 4.9 = 15.1 m
18.
Given that :
VA = VAG = 54 km/h = 15 m/s
VB = VBG = - 90 km/h = - 25 m/s
VMA = -18 km/h = - 5 m/s
i) VBA = VB – VA = - 90 – 54 = -144 km/h (- 40 m/s )
ii) VAB = VA – VB = 144 km/h
iii) VGB = VG – VB = - VBG = 90 km/h ( 25 m/s )
iv) VMG =
=
=
=
VM – VG
VM – VA + VA – VG
VMA + VAG
-18 + 54 = 36 km/h ( 10 m/s)