Chapter (2)
Kinematics of A particle
2.1 Rectilinear Motion
2.1 A car starts from rest and with constant acceleration achieves a velocity of 15 m s when it
travels a distance of 200 m . Determine the acceleration of the car and the time required.
Solution
V  15 m s
V0  0
S  200 m
(a) The acceleration of the car
V 2  V02  2a S  S 0 
152  02  2a 200  0
a  0.5625 m s 2
225  400 a
(b) The time required
V  V0  a.t
15  0  0.5625  t
t  26.67 sec
2.2 A car starts from rest with constant acceleration of
2 m s 2 traveling
a distance of
150 m along a straight road. Determine its velocity and time to travel.
Solution
V0  0
a  2m s
2
S  150 m
(a) The car velocity
V 2  V02  2a S  S 0 
2
V 2  0  2  2150  0
V 2  600
V  24.5 m s
(b) The time to travel
V  V0  a.t
24.5  0  2  t
t  12.25 sec
1
2.3 A train starts from rest at a station and travels with a constant acceleration of 1 m s 2 .
Determine the velocity of the train when t  3 sec and the distance traveled during this time.
a  1m s2
V0  0
t  3 sec
Solution
(a) The train velocity
V  V0  a.t
V  3m s
V  0  1 3
(b) The distance traveled
S  V0 t 
1 2
at
2
1
2
S  0  3   1  3
2
S  4.5 m
2.4 A car traveling with constant acceleration of 6000 km h 2 along a straight road with an initial
speed of 70 km h . How long will it take to reach a speed of 120 km h ? Also, what distance
does the car travel during this time?
Solution
a  6000 km h 2  0.463 m s 2
V0  70 km h  19.44 m s
V  120 km h  33.33 m s
(a) The time to travel
V  V0  a.t
33.33  19.44  0.463  t
t  30 sec
(b) The distance of traveling
V 2  V02  2a S  S 0 
33.332  19.44 2  2  0.463S  0
732.975  0.926 S
S  791.55 m
2
2.5 The position of a particle is given by S  8  3t 2  5t 3 m  , where t is in seconds, determine:
(a) The position of a particle at t  1 sec .
(b) The velocity of the particle at t  1 sec .
(c) The acceleration of the particle at t  1 sec .
Solution
S  8  3t 2  5t 3
(a) The position of a particle at t  1 sec
S  8  3t 2  5t 3
At t  1 sec :
2
3
S  8  31  51
S  10 m
(b) The velocity of the particle at t  1 sec
V 
dS
dt
V  0  6t  15t 2
V  6t  15t 2
At t  1 sec :
2
V  6  1  15 1
V 9 m s
(c) The acceleration of the particle at t  1 sec
a
dV
dt
but
V  6t  15t 2
a  6  30t
At t  1 sec :
a  24 m s 2
a  6  30  1
3
2.6 The position of a particle is given by S  2t 2  8t  6 m  , where t is in seconds, determine:
(a) The velocity of the particle at t  3 sec
(b) The acceleration of the particle at t  3 sec
(c) The time when the velocity of the particle become zero
Solution
S  2t 2  8t  6
(a) The velocity of the particle at t  3 sec
V 
dS
dt
V  4t  8
At t  3 sec :
V  4m s
V  438
(b) The acceleration of the particle at t  3 sec
a
dV
dt
a4
At t  3 sec :
a  4 m s2
(c) The time when the velocity of the particle become zero
When the velocity of the particle become zero
V 0
4t  8  0
4t  8
t  2 sec
4
2.7 The car in figure moves in a straight line such that its velocity is defined by V  2.7t 2  0.6t
m s  , where
t is in seconds, determine its position and acceleration when t  3 sec . When
t  0 , s  0.
Solution
V  2.7t 2  0.6t
(a) The car position at t=3 sec
t
S  So   Vdt
0
t
S  0   2.7t 2  0.6t dt
0
t
 t3
t2 
S  2.7  0.6 
3
2 0



S  0.9t 3  0.3t 2  0
S  0.9t 3  0.3t 2
3
2
S  0.93  0.33
S  24.3  2.7
S  27 m
(b) The car acceleration at t=3 sec
a
dV
 5.4t  0.6
dt
a  5.4t  0.6
a  16.8 m s 2
a  5.4  3  0.6
5
2.8 A particle travels along a straight line with a velocity of V  4t  3t 2 m s  , where t is in
seconds. Determine its position and acceleration when t  4 sec . When t  0 , s  0 .
Solution
V  4t  3t 2
(a) The car position at t=4 sec
t
S  So   Vdt
0
t
S  0   4t  3t 2 dt
0
t
 t2
t3 
S  4  3 
3 0
 2


S  2t 2  t 3  0
S  2t 2  t 3
2
3
S  24   4 
S  32  64
S  32 m
(b) The car acceleration at t=4 sec
a
dV
 4  6t
dt
a  4  6t
a  20 m s 2
a  464
6
2.9 A particle travels along a straight line with a velocity V  12  3t 2 m s  , where t is in
seconds. When t  1 sec , the particle is located 10 m to the left of the origin. Determine the
acceleration when t  4 sec , the position of the displacement of the particle at t  10 sec
Solution
V  12  3t 2
(a) The car acceleration at t=4 sec
a
dV
 0  6t
dt
a  6t
a  24 m s 2
a  6  4
(b) The car position at t=10 sec
t
S  So   Vdt
1
t
S  10   12  3t 2 dt
1
t

t3 
S  10  12t  3 
3 1



S  10  12t  t 3  12  1
S  10  12t  t 3  11
S  12t  t 3  21
3
S  12  10  10  21
S  901 m
7
2.10 Starting from rest, a particle moving in a straight line has an acceleration of
a  2t  6 m s 2  , where t is in seconds. What is the particle’s velocity when t  6 sec , and
what is its position when t  11 sec ?
Solution
a  2t  6
(a) The particle Velocity
t
V  Vo   adt
0
t
V  0   2t  6dt
0

V  t

 6t   0  0
V  t 2  6t
2
t
0
V  t 2  6t
when t=6 sec:
2
V  t 2  6t  6  6  6
V 0
(b) The particle displacement at t=11 sec
t
S  So   Vdt
0
t
S  0   t 2  6t dt
0
t
t3
t2 
S   6 
2 0
3
t3

S    3t 2   0  0
3

t3
S   3t 2
3
when t=11 sec: S 
113  3  112
S  80.7 m
3
8
2.11 A particle has an initial speed of 27 m s . If it experiences a deceleration of a  6t m s 2  ,
where t is in seconds, determine its acceleration and displacement traveled when it goes to
rest.
Solution
a  6t
(a) The particle Velocity
t
V  Vo   adt
0
t
V  27    6t dt
0
t
 t2 
V  27    6 
2 0

V  27  3t 2
The particle goes to rest: V  0
0  27  3t 2
3t 2  27
t  3 sec
(b) The particle acceleration at t=3 sec
a  18 m s 2
a  6t  6  3
(c) The particle displacement at t=3 sec
t
S  So   Vdt
0
t
S  0   27  3t 2 dt
0
t

t3 
S  27t  3 
3 0

S  27t  t 3
3
S  27  3  3
S  54 m
9
2.12 A particle moves along horizontal path with a velocity of V  3t 2  6t  m/s . If it is
initially located at the origin O. Determine:
(a) The position of the particle after 3.5 second
(b) The total distance traveled after 3.5 second
(c) The particle average velocity and average speed during the time interval.
Solution
(a) The position of the particle after 3.5 second
t
S  So   Vdt
0
t
S  0   3t 2  6t dt
0

S  t 3  3t 2

t
0
S  t 3  3t 2
At t  3.5 sec : S  3.53  33.52
S  6.125 m
(b) The total distance after 3.5 second
S  t 3  3t 2
First, determine the path motion as follows:
For V  0
0  3t 2  6t
t  2 sec
3
2
S after 2 sec  2   32  4 m
3
2
S after 3.5 sec  3.5  33.5  6.125 m
The total distance travelled= 4   4   6.125  14.125 m
(c) The particle average velocity and average speed during the time interval
average velocity 
average speed 
14.125
 4m/s
3.5
6.125
 1.75 m / s
3.5
10
2.13 A particle moves along horizontal path with a acceleration of a  2t  9  m/s2 . When
t  0 , S  1 m and V  10 m s . Determine:
(a) The velocity after 9 second.
(b) The position of the particle after 9 second.
(c) The total distance traveled after 9 second.
Solution
(a) The velocity after 9 second
t
V  Vo   adt
0
t
V  10   2t  9 dt
0


t
V  10  t 2  9t 0
V  t 2  9t  10
At t  9 sec : V  9 2  9  9  10
V  10 m s
(b) The position of the particle after 9 second
t
S  So   Vdt
0
t
S  1   t 2  9t  10dt
0
t
9
1

S  1   t 3  t 2  10t 
2
3
0
S
At t  9 sec :
1 3 9 2
t  t  10t  1
3
2
S
1 3 9 2
9   9   109   1
3
2
(c) The total distance traveled after 9 second
First, determine the path motion as follows:
For V  0
0  t 2  9t  10
t  1.3 sec
and
t  7.7 sec
11
S  30.5 m
At t  1.3 sec : S1.3 
1
1.33  9 1.32  101.3  1  7.13 m
3
2
At t  7.7 sec : S 7.7 
1
7.73  9 7.7 2  107.7   1  36.63 m
3
2
At t  9 sec : S9 
1 3 9 2
9   9   109   1  30.5 m
3
2
-30.5 m
The origin
t= 9 s
S0 =1 m
t=0 s
7.13m
t=7.7 s
-36.63 m
t=1.3 s
The total distance travelled= 7.13  1  7.13  36.63  36.63  30.5  56.02 m
12
2.14 When a train is traveling along a straight track with initial velocity of 2 m s , it begins to
accelerate at a  60 V 4 where a is expressed in m s 2 and V in m s . Determine:
(a) The train velocity after 3 sec
(b) The train position after 3 sec
Solution
(a) The train velocity after 3 sec
a
dV
 60V 4
dt
dV
 60dt
V 4
V 4 dV  60dt
t
V
4
 V dV   60dt
V0
0
t
V
V
4
dV   60dt
2
0
V
V 5 
t
 5   60t 0
 2
1 5
5
V  2   60t  0
5


V 5  32  300t
V 5  300t  32
15
V  300t  32
at t=3 sec:
15
V  300  3  32 
V  3.93 m s
(b) The train position after 3 sec
t
S  So   Vdt
0
t
15
S  0   300t  32  dt
0
13
t
 300t  32 6 5 
S

 6 5  300  0
at t=3 sec:
S
5
300t  326 5  300  0  32 6 5
6  300
S
1
300t  326 5  64
360
S
1
300  3  326 5  64
360






S  9.985 m
14
2.15 A particle travels in a straight line such that its motion described by a  4 V where a is
expressed in m s 2 and V in m s . If V  6 m s when t  2 sec , find the velocity and
acceleration of the particle at t  3 sec .
Solution
(a) The particle velocity
a
dV 4

dt V
dV 4

dt V
VdV  4dt
V
t
 V dV   4 dt
6
2
V
V 2 
t
 2   4t 2
 6
1 2
2
V  6   4t  2
2


V 2  36  8t  16
V 2  8t  20
at t=3 sec:
V 2  8  3  20
V 2  44
V  6.63 m s
(b) The particle acceleration
at t=3 sec:
a
4
V
a
4
6.63
a  0.61 m s 2
15
2.16 The acceleration of a rocket traveling upward is given by: a  6  0.02 S m s 2  , where
a is expressed in m s 2 and S in m . Determine the velocity of the rocket when S  2000 m .
Initially V  0 and S  0 when t  0
Solution
a V
dV
dS
6  0.02 S  V
dV
dS
6  0.02 S dS  VdV
S
V
 6  0.02 S dS   VdV
0
0
V
6 S  0.01 S 
2 S
0
V 2 
 
 2 0
V2
6 S  0.01 S 
2
2
V 2  12 S  0.02 S 2
when S  2000 m :
V 2  12 S  0.02 S 2
2
V 2  12  2000  0.02  2000
V 2  104000
V  322.5 m s
16
2.17 The acceleration of a particle is defined by the relation a  0.8V where a is expressed in
m s 2 and V in m s . Knowing that at t  0 the velocity is 20 m s , determine:
(a) The distance the particle with respect to the time
(b) The time and distance required for the particle to be reduced by 50% of its initial velocity
value.
Solution
a
dV
 0.8V
dt
dV
 0.8dt
V
t
V
dV
V V  0  0.8dt
0
V
t
dV
20 V  0  0.8dt
ln V V20  0.8t 0t
ln V  ln(20)  0.8t  0
V 
ln   0.8 t
 20 
V
 e - 0 .8t
20
V  20e - 0 .8t
(a) The distance the particle with respect to the time
t
S  So   Vdt
0
t
S  0   20e 0.8t dt
0
t
 e 0.8t 
S  20

  0.8  0
S
20 0.8t
e
 e0
 0.8





S  25 e 0.8t  1
S  25 1  e 0.8t
17

(b) The time and distance for the particle to be reduced by 50% of its initial velocity value.
V  0.5V0  0.5  20  10 m s
(i) Time
V  20e - 0 .8t
10  20e - 0 .8t
e - 0 .8t  0.5
t  0.866 sec
(ii) Distance



S  251  e

S  251  e
S  25 1  e 0.8t
0.80.866
0.80.866
S  12.5 m
18
2.18 The acceleration of a particle traveling along a straight line is a  8  2 S m s 2  , where S
is in meters. If V  0 and S  0 when t  0 , determine the velocity of the particle at S  2 m .
Solution
a V
dV
dS
8  2S  V
dV
dS
8  2S dS  VdV
S
V
 8  2S dS   VdV
0
0
S
V

V 2 
S2 
8
S

2


2  0  2  0

8 S  S2 
V2
2
V 2  16 S  2 S 2
when S  2 m :
V 2  16 S  2 S 2
2
V 2  16  2  2  2 
V 2  32  8
V 2  24
V  4.90 m s
19
2.19 A particle is moving along a straight line such that its speed is defined as V   4S 2  m s ,
where s is in meters. If s  2 m when t  0 , determine the velocity and acceleration as
functions of time.
Solution
dS
dt
V 
 4S 2 
dS
dt
dS
 4dt
S2
S
S
t
2
dS    4dt
2
0
1
S
S 
t
  1   4t 0

2
S
1
t
    4t 0
 S 2
 1 1
t
     4t  00
S
2


1 1
  4t
S 2
1
1
 4t 
S
2
1
S
4t 
1
2
(a) The velocity as functions of time.
1

S   4t  
2

1
2
dS
1

V 
  4t    4
dt
2

V 
4
1

 4t  
2

2
(b) The acceleration as functions of time.
1

V  4 4t  
2

2
3
a
dV
1

 4  24t    4
dt
2

a
32
1

 4t  
2

20
3
2.20 A particle travels in a straight line with accelerated motion such that a  k .S , where S is
the distance from the starting point and k is proportionality constant. For S  2 m the velocity
V  4 m s , and for S  3.5 m the velocity V  10 m s . What is the distance when V  0 .
Solution
a V
dV
dS
dV
dS
 k .S  V
 k .S dS  VdV
S
V
  k.S dS   VdV
2
4
S
V

V 2 
S2 

k


2  2  2  4


k 2
1
2
2
S  2   V 2  4 
2
2

 



 k S 2  4  V 2  16
when S  3.5 m , the velocity V  10 m s :


2
2
 k 3.5  4  10  16
 8.25 k  84
k  10.18
Then  k S 2  4  V 2  16


10.18 S 2  4  V 2  16
when V  0 :

10.18S

 4  0
10.18 S 2  4  V 2  16
2
2
 16
S 2  2.428
S  1.56 m
21
2.2 Curvilinear Motion
[1] Cartesian Coordinates
2.21 The position vector of a particle is given by: r  10t 2  2t i  t 3  5 j  0 k , determine:
(a) The particle velocity at t  2 sec .
(b) The particle acceleration at t  2 sec .
Solution
r  10t 2  2t i  t 3  5 j  0  k
(a) The particle velocity at t  2 sec
 dr
V 
dt
V  20t  2 i  3t 2  0 j
V  20t  2 i  3t 2  j
At t  2 sec : V  20  2  2 i  32 2  j
V  42i  12 j
V 
422  122
V  43.7 m s
(b) The particle acceleration at t  2 sec

 dV
a
dt
a  20i  6t  j
At t  2 sec :
a  20 i  6  2  j
a  20 i  12  j
a
202  12 2
a  23.3 m s 2
22

2.22 The velocity vector of a particle is given by: V  12t 2 i  16t 3  j  5 k m/s. The particle

starts at the position ro  0 i  4 j  3 k m, Find: (a) The acceleration vector at t  2 sec .
(b) The position vector at t  2 sec .
(c) The displacement vector from t  0 to t  2 sec
Solution
V  12t 2 i  16t 3 j  5 k
(a)The acceleration vector
a  24t i  48t 2 j  0 k

2

a  24  2  i  48  2  j  0 k
At t  2 sec :
a  48 i  192 j  0 k
(b) The position vector
t

 
r  ro   Vdt
0
t
 
r  ro   12t 2 i  16t 3 j  5 k dt


0

r  0 i  4 j  3 k  4t 3 i  4t 4 j  5t k


t
0

r  0 i  4 j  3 k  4t 3 i  4t 4 j  5t k

r  4t 3 i  4t 4  4  j  5t  3 k
At t  2 sec :

3
4
r  42  i  42   4 j  52   3 k



r  32i  68 j  13 k
(c) The displacement vector from t  0 to t  2 sec
 

S  r t  2  r t 0

S  32i  68 j  13 k  0 i  4 j  3 k

S  32i  64 j  10 k

 

23

2.23 The acceleration vector of a particle is given as: a  3t 2 i  8 j  1  t k , Where t is the

time in seconds. If the particle started from rest at position ro  2 i  3 j  0 k , determine:
(a) The velocity vector of the particle at t  2 sec
(b) The position vector of the particle at t  2 sec
(c) The displacement vector of the particle from t  0 to t  2 sec
Solution

a  3t 2 i  8 j  1  t k
(a) The velocity vector of the particle
  t 
V  Vo   a dt
0
  t 
V  Vo   a dt
0
  t
V  Vo   3t 2 i  8 j  1  t k dt


0
t

3
 t2  
V  0  t i  8t j   t  k 
2  0



 t2 
V  t 3 i  8t j   t  k
2



At t  2 sec : V  2 3 i  8  2  j   2 

2 2 k
2 

V  8 i  16 j  0 k
(b) The position vector
t

 
r  ro   Vdt
0
t

 t2  
 
r  ro   t 3 i  8t j   t  k dt
2 

0 
t
t4
 t2 1 t3  

k 
r  2 i  3 j  0 k   i  4t 2 j   
 2 2 3  0
4
24
t4
 t2 t3  

r  2 i  3 j  0 k   i  4t 2 j    k 
2 6 
4

 t2 t3 
  t4
r    2 i  4t 2  3 j    k
4

2 6

1
4
 t2 t3 
 k
2 6


At t  2 sec : r   t 4  2 i  4t 2  3 j  

4

 2 2 2 3 
  2 
2
k
r  
 2 i  42   3 j  

6 
 4

 2



2
r  6i  19 j  k
3
(c) The displacement vector of the particle from t  0 to t  2 sec
 

S  r t  2  r t 0
 
2 
S   6i  19 j  k   2 i  3 j  0 k
3 




2
S  4i  16 j  k
3
25
2.24 The position vector of a particle moving in a plane is: r  8t  4 sin 2t i  ( 4  4 cos 2t ) j .
Find:
(a) The velocity vector at any time
(b) The acceleration vector at any time
(c) Show that: V 2  2.a. y
Solution
r  8t  4 sin 2t i  ( 4  4 cos 2t ) j
(a) The velocity vector at any time
 dr
V 
dt
V  8  8 cos 2t i  (0  8 sin 2t ) j
V  8  8 cos 2t i  (8 sin 2t ) j
(b) The acceleration vector at any time

 dV
a
dt
a  16 sin 2t i  (16 cos 2t ) j
(c) Show that: V 2  2.a. y
(1) L.H .S  V 2  (8  8 cos 2t ) 2  (8 sin 2t ) 2  128(1  cos 2t )
a  (16 sin 2t ) 2  (16 cos 2t ) 2  16
y  41  cos 2t 
(2) RHS  2.a. y
RHS  2  16  41  cos 2t 
RHS  1281  cos 2t 
Then: L.H .S  R.H .S
V 2  2.a. y
26
2.25 A particle moves along the path: r  4t 4 i  3t  8t 3  j . Find:
(a) The magnitude of the particle velocity, when t= 2 sec
(b) The magnitude of the particle acceleration, when t= 2 sec
(c) The equation of the path
Solution
(a) The magnitude of the particle velocity
 dr
V 
dt
V  16t 3  i  3  24t 2  j
when t= 2 sec:

3
 
2

V  162  i  3  242  j
V  128 i  99 j
1282  992
V 
V  161.82 m s
(a) The magnitude of the particle acceleration

 dV
a
dt
a  48t 2  i  48t  j
when t= 2 sec:

2

a  482  i  48  2  j
a  192 i  96 j
a
1922  962
a  214.66 m s 2
(c) The equation of the path
r  4t 4 i  3t  8t 3  j
x  4t
4
where
x  4t 4
 x
t 
4
then
14
x
 8 
4
y  3t  8t 3
14
y  3t  8t 3
 x
y  3 
4
and
34
27
in y equation
2.26 The acceleration of a moving particle is: a  20 cos 2t i  20 sin 2t  j . The particle starts
the motion at point (8 m, 4 m) , with a velocity V0  10 m s , in the positive direction on the y–
axis . Find:
(a) The position vector equation
(b) The Cartesian equation of the path
(c) Prove that the magnitudes of acceleration and velocity are constants.
Solution
r0  8i  4 j
V0  10 j
The velocity vector:
  t 
V  Vo   a dt
0
  t
V  Vo    20 cos 2t i  20 sin 2t  j dt


0

t
V  10 j   10 sin 2t i  10 cos 2t  j 0

V  10 j   10 sin 2t i  10 cos 2t  j  0i  10 j

V  10 j   10 sin 2t i  10 cos 2t  j  10 j

V  10 sin 2t i  10 cos 2t  j




 


(a) The position vector
t

 
r  ro   Vdt
0
t

r  8 i  4 j    10 sin 2t i  10 cos 2t  j dt


0
t

r  8 i  4 j  5 cos 2t i  5 sin 2t  j o



r  8 i  4 j  5 cos 2t i  5 sin 2t  j   5i  0 j 

r  3 i  4 j  5 cos 2t i  5 sin 2t  j 

r  3  5 cos 2t  i  4  5 sin 2t  j
(b) The Cartesian equation of the path

r  3  5 cos 2t  i  4  5 sin 2t  j
x  3  5 cos 2t
and
y  4  5 sin 2t
28
From x:
x  3  5 cos 2t
cos 2t 
From y:
y  4  5 sin 2t
sin 2t 
But:
x3
5
y4
5
sin 2t 2  cos 2t 2  1
2
2
 y  4   x  3

 
 1
 5   5 
 y  42  x  32  25
(c) Prove that the magnitudes of acceleration and velocity are constants
(1) Acceleration: a  20 cos 2t i  20 sin 2t  j
a
20 cos 2t 2  20 sin 2t 2
2
(2) Velocity:
2
a  20 cos 2t   sin 2t 

V  10 sin 2t i  10 cos 2t  j
V 
a  20 m s 2
10 sin 2t 2  10 cos 2t 2
2
2
V  10 sin 2t   cos 2t 
V  10 m s
29
2.27 The acceleration of a moving particle is: a  20 sin 2t i  20 cos 2t  j . The particle starts
the motion at point (0 m, 5 m) , with a velocity V0  10 m s , in the positive direction on the x–
axis . Find:
(a) The position vector equation
(b) The Cartesian equation of the path
(c) Prove that the magnitudes of acceleration, velocity and position are constants
Solution
r0  0 i  5 j
V0  10 i
The velocity vector:
  t 
V  Vo   a dt
0
 
V  Vo    20 sin 2t i  20 cos 2t  j dt
t


0

t
V  10 i  10 cos 2t i  10 sin 2t  j 0

V  10 i  10 cos 2t i  10 sin 2t  j  10 i  0 j

V  10 i  10 cos 2t i  10 sin 2t  j  10 i

V  10 cos 2t i  10 sin 2t  j



 

(a) The position vector equation
t

 
r  ro   Vdt
0
t

r  0 i  5 j   10 cos 2t i  10 sin 2t  j dt


0
t

r  5 j  5 sin 2t i  5 cos 2t  j o



r  5 j  5 sin 2t i  5 cos 2t  j   0  5 j 

r  5 j  5 sin 2t i  5 cos 2t  j   5 j

r  5 sin 2t i  5 cos 2t  j
(3) The Cartesian equation of the path

r  5 sin 2t i  5 cos 2t  j
x  5 sin 2t
and
y  5 cos 2t
30
From x:
x  5 sin 2t
sin 2t 
x
5
From y:
y  5 cos 2t
cos 2t 
y
5
But:
sin 2t 2  cos 2t 2  1
2
2
 x  y
    1
5  5
x 2   y 2  25
(c) The magnitudes of acceleration, velocity and displacement are constants
(1) Acceleration: a  20 sin 2t i  20 cos 2t  j
a
20 sin 2t 2  20 cos 2t 2
2
2
a  20 sin 2t   cos 2t 

V  10 cos 2t i  10 sin 2t  j
(2) Velocity:
V 
a  20 m s 2
10 cos 2t 2  10 sin 2t 2
2
2
V  10 cos 2t   sin 2t 
V  10 m s

(3) Position: r  5 sin 2t i  5 cos 2t  j
r
5 sin 2t 2  5 cos 2t 2
2
2
r  5 sin 2t   cos 2t 
r  5m
31
2.28 A particle travel along the path y  0.5 x 2 , where x and y are in meters. If the particle's
component of velocity in the x direction is always Vx  5t m s , When t  0 the particle is at
point 0, 0 . Determine:
(a) How far from the origin is the particle when t  1 sec .
(b) The magnitude of the particle's velocity when t  1 sec .
(c) The magnitude of the particle's acceleration when t  1 sec .
Solution
Path equation: y  0.5 x 2
Vx 
r0  0i  0 j
Vx  5t
dx
 5t
dt
dx  5t dt
x
t
 dx   5t dt
x0
0
x
t
 dx   5t dt
0
0
x 0x   5 t 2 
2
x0 
t
0
5 2
t  0
2
x
5 2
t
2
From the path equation: y  0.5 x 2
5 
y  0.5 t 2 
2 
2
y  3.125 t 4
(a) How far from the origin is the particle when t  1 sec
The position vector:
rxi y j
r  2.5t 2  i  3.125t 4  j
when t= 1 sec:

2
 
4
r  2.51 i  3.1251
j
r  2.5 i  3.125 j
r
2.52  3.1252
r  4m
32
(b) The magnitude of the particle's velocity when t  1 sec
r  2.5t 2  i  3.125t 4  j
V 
dr
dt
V  5t  i  12.5t 3  j


when t= 1 sec: V  5  1 i  12.5  13 j
V  5 i  12.5 j
V 
52  12.52
V  13.46 m s
(c) The magnitude of the particle's acceleration when t  1 sec
V  5t  i  12.5t 3  j
a
d V
dt
a  5 i  37.5t 2  j


when t= 1 sec: a  5 i  37.5  12 j
a  5 i  37.5 j
a
52  37.52
a  37.83 m s 2
33
2.29 A particle travel along the path y  x 2 , where x and y are in meters. If the particle's
component of velocity in the y direction is always V y  3 m s , When t  0 the particle is at
point 1 m, 1 m  . Determine: (a) How far from the origin is the particle when t  2 sec .
(b) The magnitude of the particle's velocity when t  2 sec .
(c) The magnitude of the particle's acceleration when t  2 sec .
Solution
Path equation: y  x 2
Vy 
r0  i  j
Vy  3 m s
dy
3
dt
dy  3dt
y
t
 dy   3dt
y0
0
y
t
 dy   3dt
1
0
 y 1y  3t t0
y  1  3t  0
y  3t  1
From the path equation: y  x 2
3t  1  x 2
x  3t  1
(a) How far from the origin is the particle when t  2 sec .
The position vector:
rxi y j
r


3t  1 i  3t  1 j
 3  2  1 i  3  2  1 j
r   7  i  7  j
when t= 2 sec: r 
r  7  49
r  7.48 m
(b) The magnitude of the particle's velocity when t  2 sec .
12
r  3t  1
i  3t  1 j
34
V 
dr
dt
V 
3
3t  11 2 i  3 j
2
V
3
i  3 j
2 3t  1
when t= 2 sec: V 
3
i  3 j
2 3 2 1
V  0.57 i  3 j
V 
0.572  32
V  3.05 m s
(c) The magnitude of the particle's acceleration when t  2 sec
V 
3
3t  11 2 i  3 j
2
a
d V
dt
a
3 3
3t  13 2 i  0 j

2 2
a
9
3t  13 2 i
4
when t= 2 sec: a 
a
9
3  2  13 2 i
4
9
3  2  13 2 i
4
a  0.12 m s 2
a  0.12 i
35
2.30 The motion of particle A and B is described by the position vectors rA  2t i  t 2  1 j and
rB  t  2 i  2t 2  5 j . Determine:
(a) The time when the particles collide
(b) The point where the particles collide
(c) The velocity of each particle just before the collisions
Solution
(a) The time when the particles collide
For the particles collide: rA  rB
2t i  t 2  1 j  t  2i  2t 2  5 j
From x – component:
2t  t  2
t  2 sec
From y – component:
t 2  1  2t 2  5
t  2 sec
Then the two particles will collide after 2 sec
(b) The point where the particles collide
Obtain the coordinate for any particle after 2 sec
For particle (A):
x A  2t  2  2
xA  4
y A  t 2  1  2   1
2
yA  3
The point of collide is (4,3)
(c) The velocity of each particle just before the collisions
For particle (A):
rA  2t i  t 2  1 j
VA 
When t=2 sec:
d rA
 2 i  2t  j
dt
V A  2 i  2  2  j
V A  2 i  4 j
VA 
22  42
V A  4.472 m s
36
For particle (B):
rB  t  2 i  2t 2  5 j
VB 
When t=2 sec:
d rB
 1i  4t  j
dt
VB  1i  4  2 j
VB  1i  8 j
VB 
12  82
VB  8.06 m s
37
[2] Natural Coordinates
2.31 A car moves along a circular track of radius 100 m such that the distance traveled by the
car is given by the relation: S  2t 3  10t . Determine: (a) The car distance at t  2 sec
(b) The car velocity at t  2 sec
(c) The car acceleration at t  2 sec
Solution
S  2t 3  10t
(a) The car distance
At t  2 sec : S  22 3  102 
S  36 m
(b) The car velocity
V 
At t  2 sec :
dS
 6t 2  10
dt
2
V  62   10
V  34 m s
(c) The car acceleration
(i) Tangential component
dV
 12t
dt
At t  2 sec : at  12  2
at 
at  24 m s 2
(ii) Normal component
an 
V2

2

34 

100
 11.56 m / s 2
(iii) Total acceleration
a  at2  a n2
a
11.562  242
a  26.64 m / s 2
38
2.32 A car moves along a circular track of radius 100 m such that its speed is given by the
relation: V  3t  t 2  . Determine:
(a) The car acceleration at t  3 sec
(b) The car distance traveled at t  3 sec
Solution
V  3t  t 2 
(a) The car acceleration
(i) Tangential component
dV
 31  2t 
dt
At t  3 sec : at  31  2  3
at 
at  21 m s 2
(ii) Normal component
V  3t  t 2 
At t  3 sec : V  33  9  36 m / s
an 
V2

2

36

100
 12.96 m / s 2
(iii) Total acceleration
a  at2  a n2
12.962  212
a
a  24.67 m / s 2
(b) The car distance
t
S  S 0   Vdt
0
t


S  0   3t  3t 2 dt
0
t
 t2
t3 
S  3  3 
3 0
 2
At t  3 sec :
S
3 2 3
t t
2
S
3 2
3  33
2
S  40.5 m
39
2.33 A particle is traveling along a circular curve having a radius of 40 m . If it has an initial
speed of 20 m s and then begins to decrease its speed at the rate of at  0.5t m/s2,
determine:
(a) The particle velocity at t  4 sec
(b) The particle acceleration at t  4 sec
Solution
(a) The particle velocity
t
V  Vo   at dt
0
t
V  20    0.5t  dt
0

V  20  0.25t 2

t
0
V  20  0.25t 2
At t  4 sec : V  20  0.2542
V  16 m s
(b) The particle acceleration
(i) Tangential component
at  0.5t
At t  4 sec : at  0.5  4  2 m s 2
(ii) Normal component
an 
V2


162
40
 6.4 m / s 2
(iii) Total acceleration
a  at2  a n2
a
 22  6.42
 6.7 m / s 2
40
2.34 A car starts from rest on a horizontal circular curved road of 50 m radius. If the speed is
uniformly increased at a rate of 2 m s 2 , determine:
(a) The car velocity at t  4 sec .
(b) The car acceleration at t  4 sec .
(c) The car distance at t  4 sec .
Solution
at  2 m s 2
  50 m
(a) The car velocity
V  V0  at .t
V 8 m s
V  0 24
(b) The car acceleration
(i) Tangential component:
at  2 m s 2
(ii) Normal component
an 
V2

2

8

50
 1.28 m / s 2
(iii) Total acceleration
a  at2  a n2
a
1.282  22
 2.375 m / s 2
(c) The car distance
V 2  V02  2at .S
82  0  2  2  S
S  16 m
41
2.35 A bus starts from rest on a curve of 300 m radius and accelerates at constant rate of
0.75 m s 2 . Determine the velocity, distance and the time that the bus will travel before the
magnitude of its total acceleration is 0.9 m s 2 .
Solution
  300 m
at  0.75 m s 2
a  0.9 m s 2
a 2  at2  a n2
0.92  0.752  an2
a n2  0.2475
a n  0.4975 m s 2
(a) The car velocity
an 
V2

V   .a n  300  0.4975
V  12.22 m s
(b) The time
V  V0  at .t
12.22  0  0.75  t
t  16.3 sec
(c) The car distance
V 2  V02  2at .S
12.222  0  2  0.75  S
S  99.55 m
42
2.36 A motorist is traveling on a curved portion of highway of radius 350 m at a speed of
20 m s . The brakes are suddenly applied, causing the speed to decrease at a constant rate of
1.25 m s 2 . Determine the magnitude of the total acceleration of the motorist:
(a) Immediately after the brakes have been applied
(b) After time 4 sec from the brakes
Solution
  350 m
V0  20 m s
at  1.25 m s 2
(a) Immediately after the brakes have been applied
(1)
at  1.25 m s 2
(2)
V  V0  20 m s
an 
(3)
V2


202
a n  1.143 m s 2
350
a  at2  a n2
 1.252  1.1432
a
a  1.694 m s 2
(a) After time = 4 sec
(1)
at  1.25 m s 2
(2)
V0  20 m s
V  V0  at .t
V  15 m s
V  20  1.25  4
an 
(3)
V2


152
a n  0.643 m s 2
350
a  at2  a n2
a
 1.252  0.6432
a  1.41 m s 2
43
2.37 The speed of a racing car is increased at a constant rate of 100 km h to 120 km h over a
distance of 180 m along a curve of 240 m radius. Determine the magnitude of the total
acceleration of the car after it has traveled 120 m along the curve.
Solution
V0  100 km hr  27.78 m s
  240 m
V  120 km hr  33.33 m s
S  180 m
(a) Tangential component:
V 2  V02  2at .S  S 0 
33.332  27.78  2at .180  0
at  0.943 m s 2
1111.11  771.6  2  at  180
(b) Normal component
V 2  V02  2at .S  S 0 
2
V 2  27.78  2  0.943  120  0
an 
V2


V  31.6 m s
31.62
a n  4.16 m s 2
240
(c) Total acceleration
a  at2  a n2
a
0.9432  4.162
a  4.27 m s 2
44
2.38 A particle travels around a circular path having a radius of 50 m . If it is initially traveling
with a speed of 10 m s and its speed then increases at a rate of at  0.05V m s 2  , determine:
(a) The magnitude of the particle’s velocity when t  4 sec .
(b) The magnitude of the particle’s acceleration when t  4 sec .
Solution
(a) The magnitude of the particle’s velocity when t  4 sec .
at  0.05V
dV
 0.05V
dt
V
t
dV
V V  0 0.05dt
0
ln V VV
0
 0.05t 0
t
ln V  ln V0   0.05t  0
ln
V
 0.05t
V0
V
 e 0.05t
V0
V  V0 e 0.05t
At t  4 sec : V  10e 0.054
V  12.2 m s
(b) The magnitude of the particle’s acceleraton when t  4 sec .
(i) a n 
V2


12.22
a n  2.98 m s 2
50
at  0.61 m s 2
(ii) at  0.05V  0.05  12.2
(c) Total acceleration
a  at2  a n2
a
2.982  0.612
a  3.05 m s 2
45
2.39 The truck travels in a circular path having
a radius of 50 m at initial speed of 4 m s . If

its speed is increased by rate of V  0.05 S
m s  .
2
Determine
its
velocity
and
acceleration when it has moved S  10 m .
Solution

  50 m
(a) The truck velocity
V0  4 m s
V  0.05 S  a t
at  0.05 S
V
dV
 0.05 S
dS
V dV  0.05 S dS
V
S
 V dV   0.05 S dS
V0
S0
2
V
S
V 
S2 
 2   0.05 2 
 4
 0
V
2
2


 4   0.05 S 2  0

V 2  16  0.05 S 2
V 2  0.05 S 2  16
V  0.05 S 2  16
At S  10m : V  0.05 102  16
V  21
V  4.58 m s
(b) The truck Acceleration
(1) a n 
V2


4.582
50
 0.42 m s 2
(2) at  0.05 S
At S  10m : at  0.05  10  0.5 m s 2
(3)
a  at2  a n2
a
0.52  0.422
a  0.653 m s 2
46
2.40 The car passes point A with a
speed of after which its speed is
defined by V  25  0.15 S  m s .
Determine the magnitude of the
car’s acceleration when it reaches
point B , where S  51.5 m and
x  50 m .
Solution
V  25  0.15 S 
(1) The car velocity at point (B)
VB  25  0.15 S B
VB  25  0.15  51.5
VB  17.275 m s
(2) Radius of curvature
x2
625
dy
 2x
y\ 

dx 625
d2y  2
y \\  2 
dx
625
y  16 
1  y 

\2 3 2
y\\
1   2 x 625 

2 32
 2 625
at point (B): x  50 m
1   2  50 625 

2 32
 2 625

1.038645
2 625
  324.58 m
(3) Car acceleration
(i) a n 
VB2


17.2752
a n  0.92 m s 2
324.58
dV
 25  0.15 S   0  0.15
dS
at  0.15 25  0.15 S 
(ii) at  V
at point (B): S  51.5 m
at  0.15 25  0.15  51.5
at  2.59 m s 2
(iii) a  at2  a n2
a
 2.592  0.922
a  2.75 m s 2
47
2.41 Starting from rest, the motorboat travels
around the circular path,   50 m , at a
speed V  0.2t 2
m s  , where
t
is in
seconds. Determine the magnitudes of the
boat's velocity and acceleration at the
instant t  3 sec .
V  1.8 m s , a  1.2 m s 
2
2.42 A toboggan is traveling down along
a curve which can be approximated by
the parabola y  0.4 x 2 . Determine the
magnitude of its acceleration when it
reaches point A, where its speed is
V A  8 m s , and it is increasing at the
rate of 4 m s 2 .
a  8.61 m s 
2
48