Question

A motor bike starting from rest accelerates at a constant rate of 3.0ms-2
What is the velocity of the car after it travelled 45.5m

Answer 1

We know that:

$x=x_0+v_0\times t+ \dfrac{a\times t^2}{2}$

Or:

$l=x-x_0=v_0\times t+ \dfrac{a\times t^2}{2}$

For this task $v_0=0$ , so:

$l=\dfrac{a\times t^2}{2} => t=\sqrt{\dfrac{2l}{a}}=\sqrt{\dfrac{2*45.5}{3}}=\sqrt{\dfrac{91}{3}}\approx5.51 s(seconds)$

We know that:

$a=\dfrac{v-v_0}{t}=\dfrac{v}{t}=> v=a\times t= 3\times 5.51=16.53 mps$ (meters per second)