Question

A runner accelerates uniformly from a velocity of 4 m s^-1 to a velocity of 8 m s^-1 in adistance of 150 metres.a)Find the acceleration of the runner.b) Find the amount of time elapsed during the change of velocity.q

Answer 1

a. 0.32 m/s²

b. 12.5 s

Step-by-step explanation:

We know that the final velocity vf = 8 m/s, the initial velocity vi = 4 m/s and the distance traveled x = 150 m. Then, we can calculate the acceleration using the following equation

`\begin{gathered} v_f^2=v_i^2+2ax \\ \\ v_f^2-v_i^2=2ax \\ \\ (v_f^2-v_i^2)/(2x)=a \end{gathered}`

Replacing the values, we get

`\begin{gathered} \frac{8^2-4^2}{150\text{ }}=a \\ \\ (64-16)/(150)=a \\ \\ 0.32\text{ m/s}^2=a \end{gathered}`

Therefore, the acceleration of the runner was 0.32 m/s²

Now, the amount of time elapsed is equal to

`\begin{gathered} a=(v_f-v_i)/(t) \\ \\ t=(v_f-v_i)/(a) \\ \\ t=(8-4)/(0.32) \\ \\ t=12.5\text{ s} \end{gathered}`

Therefore, the amount of time elapsed during the change of velocity was 12.5 seconds.