Question

1. a soccer ball rolls 110 m to a stop from a speed of 21 m/s. what was the balls acceleration?

Explanation:

Step1: Identify the kinematic equation

We know the initial velocity \( u = 21\space m/s \), final velocity \( v = 0\space m/s \) (since it stops), and displacement \( s = 110\space m \). The kinematic equation that relates \( u \), \( v \), \( a \), and \( s \) is \( v^{2}=u^{2}+2as \).

Step2: Rearrange the equation to solve for \( a \)

From \( v^{2}=u^{2}+2as \), we can rearrange it to \( a=\frac{v^{2}-u^{2}}{2s} \).

Step3: Substitute the values

Substitute \( v = 0 \), \( u = 21 \), and \( s = 110 \) into the equation:
\( a=\frac{0^{2}-21^{2}}{2\times110}=\frac{- 441}{220}\approx - 2.0045\space m/s^{2} \) (the negative sign indicates deceleration)

Answer:

The acceleration of the soccer ball is approximately \(-2.0\space m/s^{2}\) (or \(2.0\space m/s^{2}\) deceleration)