Chapter 18 : Motion Under Variable Acceleration „„„„„ 393

Integrating both sides of the above equation,

23

2

23

–47.5

26

tt

stC=++

3

2

–47.5 2 2

t

=++ttC...(v)

where C 2 is the second constant of integration. Now substituting the values of t = 10 and s = 85
in above equation,

3

2

22

(10)

85 (10) – 47.5 10 75

2

=+×+=+CC

∴ C 2 = 85 – 75 = 10

Substituting this value of C 2 in equation (v),

3

(^2) –4.7510
2
t
st=++t
Now for the distance from the origin at the time of start of observation , substituting t equal to 0 in the
above equation,
s = 10 m Ans.
(c) Time after start of observations in which the velocity becomes zero
Substituting the value of v equal to 0 in equation (iv),
32
0 2 – 47.5
2
t
=+t
Multiplying both sides by – 2 and rearranging
3 t^2 – 4t – 95 = 0
This is a quadratic equation in t,
∴
4 (4)^24395
6.33 s
23
t
+± +××

×
Ans.
EXERCISE 18.2

1. The motion of a body is given by an equation :
   a = t^2 – 2t + 2

where a is acceleration in m/s^2 and t is time in seconds.The velocity and displacement of

the body after 1 second was

1

6m/s

3

and

3

14 m

4

respectively. Find the velocity and

displacement after 2 seconds. (Ans.

22

7m/s;21m

33

)

2. A body starting from rest, moves along a straight line with an acceleration whose equation
   is given by :
   2
   4–
   9

t

a=

where a is in m/s^2 and t in seconds. Find (a) velocity after 6 seconds, and (b) distance

traversed in 6 seconds. (Ans.16 m/s ; 60 m )