Question

8.a squash ball with an initial velocity of 25 m/s w is hit by a squash racket, changing its velocity to 29 m/s e in 0.25 s. what is the squash ball’s average acceleration? ans: 2.2 × 102 m/s2 e

Explanation:

Step1: Define the formula for average acceleration

The formula for average acceleration \(a_{avg}=\frac{\Delta v}{\Delta t}\), where \(\Delta v = v_{f}-v_{i}\)

Step2: Assign directions as positive or negative

Let the east - ward direction be positive. Then \(v_{i}=- 25\space m/s\) (west - ward) and \(v_{f}=29\space m/s\) (east - ward), \(\Delta t = 0.25\space s\)

Step3: Calculate the change in velocity

\(\Delta v=v_{f}-v_{i}=29-(-25)=29 + 25=54\space m/s\)

Step4: Calculate the average acceleration

\(a_{avg}=\frac{\Delta v}{\Delta t}=\frac{54}{0.25}=216\space m/s^{2}\approx2.2\times10^{2}\space m/s^{2}\) (rounded to two significant figures) and the positive sign indicates the east - ward direction

Answer:

\(2.2\times 10^{2}\space m/s^{2}\space[E]\)