Question

mr.pierce accelerates from rest to 21.1 m/s on his way to first base (27.4m). what was mr.pierces acceleration?

Answer 1

We know that

• The initial speed is null because Mr. Pierce accelerates from rest.

,

• The final speed is 21.1 m/s.

,

• The distance traveled is 27.4 meters.

To find the acceleration, we use a formula that relates initial speed, final speed, acceleration, and distance.

`v^2_f=v^2_0+2ad`

Let's use the given magnitudes and solve for a.

`\begin{gathered} (21.1\cdot(m)/(s))^2=0^2+2a(27.4m) \\ 445.21\cdot(m^2)/(s^2)=54.8m\cdot a \\ a=(445.21\cdot(m^2)/(s^2))/(54.8m) \\ a\approx8.12\cdot(m)/(s^2) \end{gathered}`

Therefore, Mr. Pierce's acceleration is 8.12 m/s^2.