Question

20. The position of a particle moving in a straight line is given by y=38 +1 +5 (where y is in cm and tis in second),
then find the acceleration of the particle at time t = 2 s.
(1) 0.16 m/s2
(2) 0.32 m/s2
(3) 0.38 m/s2
(4) 0.61 m/s2​

Answer 1

Explanation:

The relation is

dt

dx

=v where x is displacement and v is velocity which is given as v=3−t

so we have dx=vdt

after 3sec the direction of motion will get reversed.

so

on integrating both sides we get ∫

X=0

X

dx=∫

t=0

t=3

(3−t)dt=9−

2

9

=4.5meter

now integrating both sides for t=3 to t=5

we have the displacement as

∫

X=4.5

X

dx=∫

t=3

t=5

(3−t)dt=−2meter

so the net distance will be X=(2)+4.5=6.5meter