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Question

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v0. The distance travelled by the particle in time t will be
  1. v0t+13bt2
  2. v0t+13bt3
  3. v0t+12bt2
  4. v0t+16bt3

A
v0t+13bt2
B
v0t+16bt3
C
v0t+13bt3
D
v0t+12bt2
Solution
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here it is given that a=bt
wkt, a=change in velocitychange in time=dvdt
dvdt=bt now integrating this we will get v
v=bt22+c
At t=0 s, v=v0, Thus v0=c
v=bt22+v0
Now. v=change in positionchange in time=dsdt=bt22+v0
ds=(bt22+v0)dt
Integrating the above we get, s=bt36+v0t+cat t=0, s=0c=0
i.e. s=bt36+v0t

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