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# Answer to Question #114716 in Mechanics | Relativity for Abie Sanneh

Question #114716
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
1
2020-05-08T16:06:19-0400

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)

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