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

What is the velocity of the car after it travelled 45.5m

Expert's answer

*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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