1. SOLVED PROBLEMS
1. Position of a particle moving along x-axis is given by 3
2
t
t
4
t
3
x 

 , where x is in meters and t in
seconds.
(a) Find the position of the particle at s
2
t  .
(b) Find the displacement of the particle in the time interval from 0
t  to s
4
t  .
(c) Find the average velocity of the particle in the time interval from s
2
t  to s
4
t  .
(d) Find the velocity of the particle at s
2
t  .
Solution:
(a) 3
2
)
t
( t
t
4
t
3
x 


 2 3
(2)
x 3 2 4 (2) (2) 6 4 4 8 2m
           .
(b) 0
x )
0
( 
m
12
)
4
(
)
4
(
4
4
3
x 3
2
)
4
( 




 .
Displacement m
12
x
x )
0
(
)
4
( 


(c) s
/
m
7
s
/
m
2
)
2
(
12
)
2
4
(
x
x
v
)
2
(
)
4
(








(d)
2
t
3
t
8
3
dt
dx



 s
/
m
1
)
2
(
3
2
8
3
dt
dx
v 2
)
2
(
)
2
( 













2. A body moving in a curved path possesses a velocity 3 m/s towards north at any instant of its motion.
After 10s, the velocity of the body was found to be 4 m/s towards west. Calculate the average accel-
eration during this interval.
Solution:
To solve this problem the vector nature of velocity must be taken into account. In the figure, the initial
velocity 0
v and the final velocity v are drawn from a common origin. The vector difference of them is
found by the parallelogram method.
The magnitude of difference is
2
2
0 AC
OA
OC
v
v 



s
/
m
5
3
4 2
2



The direction is given by
75
.
0
4
3
tan 

 , º
37


N
B
D
A
S
E
O
W
C

3 m/s
4 m/s
0
v

v
0
v
v 
0
v
A
 Average acceleration
0
| v v | 5
t 10

 
 
= 0.5 m/s2
at 37º South of West.
10 20 30 40 50 60 70 80
Velocity(in
m/s.)
time(in sec.)
20
40
60
80
3. The velocity time graph of a moving object is given in
the figure. Find the maximum acceleration of the body
and distance travelled by the body in the interval of
time in which this acceleration exists.
Solution:

2. Acceleration is maximum when slope is maximum.
2
max s
/
m
6
30
40
20
80
a 



Displacement is the area under the curve.

1
S (20 80)(10) 500m
2
  
4. A monkey is climbing a vertical tree with a velocity of 10 m/s while a dog runs towards the tree chasing
the monkey with a velocity of 15 m/s. Find the velocity of the dog relative to the monkey.
Solution:
Velocities are shown in the figure
Velocity of the dog relative to the monkey
= velocity of Dog – velocity of monkey
M
D V
V




)
V
(
V M
D





This velocity is directed along OM and its magnitude is
M
V

D
V M
V

M
V

O D

M
D
V
18
100
225
10
15 2
2



 m/s
This velocity makes an angle  with the horizontal, where
3
2
15
10
tan 

 or
º
7
.
33
3
2
tan 1








 
5. Asteel ball is dropped from the roof of a building. An observer standing in front of a window 1.5m high
notes that the ball takes
10
1
s to fall from the top to the bottom of the window. The ball reappears at the
bottom of the window 2s after passing it on the way down. If the collision between the ball and the
ground is perfectly elastic, then find the height of the building. Take g = 10 m/s2
.
Solution:
Since collision is perfectly elastic, the speed of the ball just before collision is equal to the speed of the
ball just after collision. Hence time of descent is equal to the time of ascent. Therefore time taken by the
ball to reach the ground from the bottom of the window is 1 sec.
Let u be the speed of the ball when it is at the top of the window

100
1
10
2
1
10
1
u
5
.
1 


 . 






 2
t
a
2
1
t
u
s



 u = 14.5 m/s
 ball is dropped hence its initial speed is 0
It t be the time taken by the ball to acquire the speed of 14.5 m/s, then
H 1.5 m
14.5 0 10 t, (v u at)
    

 
 t = 1.45 sec
Hence total time of descent is given by
s
55
.
2
1
10
1
45
.
1
T 



If H be the height of the building,
then
2
gt
2
1
0
H 


3. 
2
)
55
.
2
(
10
2
1
H 


 5
.
32
H  m
6. A block of ice starts sliding down from the top of an inclined roof of a house along a line of the greatest
slope. The inclination of the roof with the horizontal is 30º. The heights of the highest and lowest points
of the roof are 8.1 m and 5.6 m respectively. At what horizontal distance from the lowest point will the
block hit the ground? Neglect any friction. [g = 9.8 m/s2
]
Solution:
Acceleration of the block along the greatest slope is equal to a = g sin 30º
Distance travelled by the block along the greatest slope is equal to
m
5
º
30
sin
)
6
.
5
1
.
8
(
S 

 .
If u be the speed of the block when it is just about to leave the roof then
5
º
30
sin
g
2
0
u2



u = 7 m/s
If t be the time taken to hit the ground then
2
2
t
8
.
9
2
1
t
2
7
gt
2
1
t
º
30
sin
u
6
.
5 




5.6
8.1
u sin30º
u cos30º
g cos30º g g sin30º
30º
u
 0
8
t
5
t
7 2



7
2
)
8
(
)
7
)(
4
(
25
5
t







 s
14
78
.
15
5
t


 , -ve value is to be rejected.
i.e., 77
.
0
14
78
.
15
5
t 


 sec.
Horizontal distance travelled is equalto
m
14
78
.
10
2
3
7
t
º
30
cos
u
x 


 67
.
4
x  m
7. A man wants to cross a river 500 m wide. The rowing speed of the man relative to water is 3 km/hr and
the river flows at the speed of 2 km/hr. If the man’s walking speed on the shore is 5 km/hr, then in which
direction should he start rowing in order to reach the directly opposite point on the other bank in the
shortest time.
Solution:
Lets 
m
v

velocity of the man relative to ground.
v

= velocity of the man relative to water
u

= velocity of water,
,
given 1
| u | 2kmh


, 1
| v| 3kmh


Let him start at an angle  with the normal v
A
B
x
y
 u
Net velcoity of man m
v u v
 
  
ĵ
cos
v
î
)
sin
v
u
(
vm 





Hence time taken by the man to cross the river is


cos
v
5
.
0
t1
 Drift of the man along the river is

4. 1
t
)
sin
v
u
(
x 


)
sin
v
u
(
x 



cos
v
5
.
0
Time taken by the man to cover this distance is



















 tan
sec
v
u
1
.
0
5
tan
v
sec
u
5
.
0
t2
Therefore, total time 2
1 t
t
T 

 




 tan
1
.
0
sec
v
u
1
.
0
sec
v
5
.
0
T
Putting the value of u and v, we get






 tan
1
.
0
sec
3
2
1
.
0
sec
3
5
.
0
T



 tan
1
.
0
sec
3
7
.
0
 



tan
sec
3
7
.
0
d
dT
– 0.1 
2
sec
for T to be minimum , 0
d
dT


 )
7
/
3
(
sin 

 )
7
/
3
(
sin 1



8. Two particles A and B move with constant velocities 1
v and 2
v along two mutually perpendicular
straight lines towards the intersection point O. At moment t = 0, the particle were located at distance 1
l
and 2
l from O, respectively. Find the time, when they are nearest and also the shortest distance
between them.
Solution:
Method I:
 ĵ
v
î
v
v
v
v 2
1
B
A
AB 






Minimum distance is the length of the perpendicular to B
A
v

from B.
If  is the angle between the x – axis and AB
v

, then
1
2
1
2
v
v
v
v
tan 



In 1
1
2
l
v
v
tan
OA
OD
,
AOD 



Therefore
1
1
2
2
1
2
v
l
v
l
v
OD
l
BD




A

v1
y
O x
v2
D
C
B AB
v
 l1
l2
In ,
BCD

BD
BC
cos 


5.  2
2
2
1
1
1
1
2
2
1
v
v
v
v
l
v
l
v
cos
BD
BC






 2
2
2
1
1
2
2
1
v
v
|
l
v
l
v
|
BC



The required time
AB
AB v
DC
AD
v
AC
t 





2 2
1 1 2 2 1 2
1 2 2 2
1 1
1 2
1 1 1 2 2
2 2
2 2 2 2
1 2
1 2 1 2
v v v
v v
v v
v v
sec BCtan v v
v v
v v v v
l l l
l l l

  

   
 

 
.
Method II:
After time ‘t’, the position of the point A and B are )
t
v
l
( 1
1  and )
t
v
l
( 2
2  , respectively.
.
The distance L between the points A’ and B’ are
2
2
2
2
1
1
2
)
t
v
l
(
)
t
v
l
(
L 


 ...(i)
Differentiating with respect to time,
)
v
)(
t
v
l
(
2
)
v
)(
t
v
l
(
2
dt
dL
L
2 2
2
2
1
1
1 




 From minimum value of 0
dt
dL
,
L 
2
2
1
1
2
2
2
1 v
l
v
l
t
)
v
v
( 

 .
Or 2
2
2
1
2
2
1
1
v
v
v
l
v
l
t



Putting the value of t in equation (1)
A
v1 A'
L
l1
O
B'
B
l2
v2
2
2
2
1
1
2
2
1
min
v
v
v
l
v
l
L


 .
9. A particle moving with uniform acceleration in a straight line covers a distance of 3 m in the 8th
second
and 5 m in the 16th
second of its motion. What is the displacement of the particle from the beginning of
the 6th
second to the end of 15th
second?
Solution:
The distance traveled during the nth second of motion of a body is given by
)
1
n
2
(
a
2
1
u
Sn 

 2
1
2
S un an
 
For the motion during the 8th
second,
2
1
( 1) ( 1)
2
u n a n
  
2
a
15
u
)
1
16
(
a
2
1
u
3 



 ...(i)
For the motion during the 16th
second,
2
a
31
u
)
1
32
(
a
2
1
u
5 




Subtracting equations (i) from (ii)
2
a
8 
or acceleration
2
ms
4
1
a 

6. From equation (i), 1
15 1 9
u 3 ms
2 4 8

 
   
 
 
Now, the velocity at the end of 5 s (velocity at the beginning of 6th
second)
a
5
u
v1 

The velocity at the end of 15th
s, a
15
u
v2 

Average velocity during this interval of 10 seconds
2
v
v 2
1 

a
10
u
2
)
a
15
u
(
)
a
5
u
(






Distance travelled during this interval
S = average velocity x time t
)
a
10
u
( 


9 10 290
10 36.25m
8 4 8
 
    
 
 
.
10. An automobile can accelerate or decelerate at a maximum value of
2
s
/
m
3
5
and can attain a maximum
speed of 90 km/hr. If it starts from rest, what is the shortest time in which can travel one kilometre, if it
is to come to rest at the end of kilometre run?
Solution:
In order that the time of motion be shortest, the car should attain the maximum velocity with the
maximum acceleration after the start, maintain the maximum velocity for as along as possible and then
decelerate with the maximum retardation possible, consistent with the condition that, the automobile
should come to rest immediately after covering a distance of 1 km.
Let 1
t be the time of acceleration, 2
t be the time of uniform velocity and 3
t be the time of retardation.
Now, maximum velocity possible = 90 km/hr
18
5
90
 m/s = 25 m/s
s
15
3
5
0
25
a
u
v
t1 




Similarly, the me of retardation is also given by
s
15
3
5
25
0
t3 



During the period of acceleration, the distance covered
= average velocity  time
25 0
15 187.5m
2

  
During the period of deacceleration, the distance covered is the same the hence
= 187.5 m
the total distance covered under constant velocity = 1000 – 375 = 625 m
Time of motion under constant velocity. s
25
25
625
t2 

the shortest time of motion 55
15
25
15
t
t
t 3
2
1 





 seconds.
11. Two particles are projected at the same instant from two points A and B on the same horizontal level
where AB = 28 m, the motion taking place in a vertical plane through AB. The particle from A has an

7. initial velocity of 39 m/s at an angle 






13
5
sin 1
with AB and the particle from B has an initial velocity
of 25 m/s at an angle 






5
3
sin 1
with BA. Show that the particle would collide in mid-air find when and
where the impact occurs.
Solution:
AB = 28m.
At A, a particle is projected with velocity u = 39 m/s. 1
u and 2
u are its horizontal and vertical compo-
nents respectively. The angle u makes with AB cos 1
 .
Given that
13
5
sin 1 
 
13
12
cos 1 
 .
Similarly for the particle projected from B, with velocity v =25 m/s, 1
v and 2
v are the horizontal and
vertical components respectively.
5
3
sin 2 
 
5
4
cos 2 
 .
u2
v2
u
v
A u1
1
28m v1
2
B
Now 15
13
5
39
sin
u
u 1
2 



 m/s
15
5
3
25
sin
v
v 2
2 



 m/s
The vertical components of the velocities are the same at the start. Subsequently at the other instant t
their vertical displacement are equal and have a value
2
t
9
.
4
t
15
h 

which means that the line joining their positions at the instant t continues to be horizontal and the
particles come closer to each other.
Their relative velocity in the horizontal direction
= 2
1 cos
25
cos
39 


s
/
m
56
20
36
5
4
25
13
12
39 






Time of collision s
5
.
0
56
28
56
AB


 , after they were projected.
Height at which the collision occurs
2
2
)
5
.
0
)(
8
.
9
(
2
1
)
5
.
0
(
15
at
2
1
ut 



= 6.275 m
The horizontal distance of the position of collision from A 1 1
(u cos )
  (time of collision)
18
s
5
.
0
13
12
39 


 m
12. A man can swim with a velocity 1
V relative to water in a river flowing with speed 2
V .Show that it will
take him 2
2
2
1
1
V
V
V

times as long to swim a certain distance upstream and back as to swim the same
distance and back perpendicular to the direction of the stream 2
1 V
V
(  ).

8. Solution:
Suppose the man swims a distance x up and the same distance down the stream.
Velocity of man upstream relative to the ground 2
1 V
V 
 .
Time taken for this,
2
1
1
V
V
x
t


Velocity of man downstream relative to the ground 2
1 V
V 

Time taken for this,
2
1
2
V
V
x
t


Total time taken 2
2
2
1
1
2
1
2
1
2
1
V
V
x
V
2
V
V
x
V
V
x
t
t







Next the man intends crossing the river perpendicular to the direction of the stream. If he wants to
cross the river straight across he must swim in a direction OM such that the vector sum of velocity of
man + velocity of river will give him a velocity relative to the ground in a direction perpendicular to the
direction of the stream. In the Figure the velocity relative to the ground is OR and the magnitude of
2
2
2
1 V
V
OR 

Now the man swims a distance x up and x down perpendicular to the river flow. Time taken for this,
2
2
2
1 V
V
x
2
t


Then the ratio,
1
2 2
1 2 1 2
2 2
1 2
2V x
t t (V V )
2x
t
v v
 


2
1 V
V 
1
V
2
V
M R

O
2
2
2
1
1
2
2
2
1
2
2
2
1
1
V
V
V
x
2
V
V
V
V
x
V
2





 .
13. A particle thrown over a triangle from one end of a horizontal base of a vertical triangle gazes the
vertex and falls on the other end of the base. If  and  be the base angle and  the angle of
projection, prove that 
tan = 

 tan
tan .
Solution:
The situation is shown in the figure.
From figure,
)
x
R
(
y
x
y
tan
tan






P(x, y)
B
x (R-x)
X
Y
u
O

 
A
where R is the range.

y(R x) xy
tan tan
x(R x)
 
   

or
)
x
R
(
R
x
y
tan
tan





 ...(i)
we known









R
x
1
tan
x
y

9. or )
x
R
(
R
x
y
tan



 ...(ii)
from equation (i) and (ii), we have




 tan
tan
tan .
14. A batsman hits a ball at a height of 1.22 m above the ground so that ball leaves the bat an angle 45º with
the horizontal. A7.31 m high wall is situated at a distance of 97.53 m from the position of the batsman.
Will the ball clear the wall if its maximum horizontal distance from point of projection is 106.68 m. Take
g = 10 m/s2
.
Solution:
 g
2
sin
v
)
range
(
R
2
0 

1.22 m
106.68m
A B
45º
v0
y
x
 Rg
2
sin
Rg
v2
0 

 as
º
45


 Rg
v0 
Equation of trajectory
º
45
cos
v
2
gx
º
45
tan
x
Y 2
2
0
2


2
1
Rg
2
gx
x
2


using (1)
Putting x = 97.53, we get
35
.
8
10
68
.
106
)
53
.
97
(
10
53
.
97
y
2





Hence height of the ball from the ground level is
h = 8.35 + 1.22 = 9.577 m.
as height of the wall is 7.31 m so the ball will clear the wall.
15. A particle projected with velocity 0
v , strikes at right angles a plane through the point of projection and
of inclination  with the horizontal. Find the height of the point struck, from horizontal plane through the
point of projection.
Solution:
Let  be the angle between the velocity of projection and the inclined plane.



 
 sin
v
v
,
cos
v
v 0
y
0
0
x
0



 sin
g
ax



 cos
g
ay
 t
sin
g
cos
v
)
t
(
v 0
x 




At the point of impact 0
vx 

 


sin
g
cos
v
t 0
...(i)
Also y at the point is zero.


y' x'
x
y
x
y
90º

10.  0
t
cos
g
2
1
t
sin
v 2
0 



 


cos
g
sin
v
2
t 0
...(ii)
From (i) and (ii)
0 0
v cos 2v sin
gsin gcos
 

 
1
tan cot
2
   ...(iii)
0
x v cos( )t
   
 








sin
g
cos
v
sin
sin
cos
cos
v 0
0  





 cos
sin
cot
cos
g
v 2
2
0
2
2
0
2 2 2
v 2 cot 2
cot
g 4 cot 4 cot 4 cot
 
  
 
 
  
 
 
     
 
 
(using
1
tan cot
2
   )



 2
2
0
cot
4
cot
2
g
v
From figure
 





 tan
cot
4
cot
2
.
g
v
tan
x
y 2
2
0

)
cot
4
(
g
v
2
y 2
2
0


 .
16. A ball is thrown from the origin in the x – y plane with velocity 28.28 m/s at an angle 45º to the x-axis.
At the same instant a trolley also starts moving with uniform velocity of 10 m/s along the positivex-axis.
Initially rear end of the trolley is located at 38m from the origin. Determine the time and position at
which the ball hits the trolley.
Solution:
Let t be the instant at which the ball hits rear face AB of the trolley.
Then 38
t
)
u
45
cos
v
( 0
0 


Or s
8
.
3
10
º
45
cos
28
.
28
38
u
º
45
cos
v
38
t
0
0





y
v0
O 38m
45º
3m
B
A
C
D
2m
x
10m/s
At s
8
.
3
t  , the y – coordinate of the ball is
2
2
0 t
5
t
20
gt
2
1
t
)
º
45
sin
v
(
y 



or m
8
.
3
)
8
.
3
(
5
)
8
.
3
(
20
y 2



Since 3.8 m > 2m, therefore, the ball cannot hit the rear face of the trolley. Now, we assume that the ball
hits the top face BC of the trolley, and let t’ be that instant. Then,
2
t
5
t
20
2
y 




or 0
4
.
0
t
4
t 2





s
9
.
3
t 

Let d be the distance from the point B at which the ball hits the trolley.
Then,
m
1
)
8
.
3
9
.
3
)(
10
20
(
)
t
t
)(
u
º
45
cos
v
(
d 0
0 








11. 17. Find the radius of a rotating wheel if the linear velocity 1
v of a point on the rim is 2.5 times greater
than the linear velocity 2
v of a point 5 cm closer to wheel axle.
Solution:
Let the radius of the disc = r (in cm.)
 



r
)
5
r
(
v
v
1
2
1
v is 2.5 times greater than 2
v
 r
5
r
v
5
.
2
v
2
2 

 5
.
12
r
5
.
2
r 

 5
.
12
r
5
.
1 
 33
.
8
5
.
1
5
.
12
r 
 cm.
18. A particle is revolving with a constant angular acceleration  in a circular path of radius r. Find the
time when the centripetal acceleration will be numerically equal to the tangential acceleration.
Solution:
Le the speed of the particle after time t from starting be v
 The centripetal acceleration
2
2
r r
r
v
a 

 & the corresponding angular speed t


 .
 2
2
2
r t
r
)
t
(
r
a 


 ...(i)
We known that the tangential acceleration 
 r
at ...(ii)
Since, t
r a
a  (given)
 

 r
t
r 2
2



1
t .