Question

A car starts from rest with acceleration bt2 (where b is dimensional constant and t is in second). If total travelled distance is 108 m after 6 s, then magnitude of b is

Answer 1

integrate dv from 0 to v and ds from 0 to s

Answer 2

Answer:

$b=1\ m/s^4$

Explanation:

Given that the particle starts from rest with a constant acceleration given by:

$a = bt^2\\\Rightarrow \dfrac{dv}{dt}=bt^2\\\Rightarrow dv=bt^2dt$

On integrating both sides, we have

$v-0= b\dfrac{t^3}{3}\\\Rightarrow v= b\dfrac{t^3}{3}\\\Rightarrow \dfrac{ds}{dt}= b\dfrac{t^3}{3}\\\Rightarrow ds=b\dfrac{t^3}{3}dt$

On integrating both sides, we have

$s-0= b\dfrac{t^4}{3\times 4}\\\Rightarrow s=b\dfrac{t^4}{12}$

At t = 6, we have

$108\ m = b\dfrac{6^4}{12}\\\Rightarrow b=\dfrac{108\ m \times 12}{6\ s^4}\\\Rightarrow b=1\ m/s^4$

Hence, the value of $b = 1\ m/s^4$.