The ball is served from rest and attains a speed of 131 mi/h ≈ 211 km/h ≈ 58.6 m/s.
a. If the racket is in contact with the ball over a distance of 0.15 m, then it applies a constant acceleration a such that
[tex]\left(58.6\dfrac{\rm m}{\rm s}\right)^2 = 2a(0.15\,\mathrm m) \implies a = \dfrac{\left(58.6\frac{\rm m}{\rm s}\right)^2}{2(0.15\,\mathrm m)} \approx \boxed{11,400\dfrac{\rm m}{\mathrm s^2}}[/tex]
b. Since acceleration is constant, we can use the definition of average acceleration:
[tex]a=a_{\rm ave} = \dfrac{\Delta v}{\Delta t}[/tex]
Solve for ∆t :
[tex]a_{\rm ave} = a = \dfrac{58.6\frac{\rm m}{\rm s}}{\Delta t} \implies \Delta t = \dfrac{58.6\frac{\rm m}{\rm s}}a \approx 0.00512 \,\mathrm s = \boxed{5.12\,\mathrm{ms}}[/tex]
