Question

The fast server in women's tennis is Venus Williams, who recorded a serve of 130 mi/h (209 km/h) in 2007. If her racket pushed on the ball for a distance of 0.15 m, what was the average acceleration of the ball during her serve?

Answer 1

The motion of the ball is an uniformly accelerated motion, so we can use the following equation to find the acceleration:

$v^2 -u^2 = 2 a S$

where

$v=209 km/h=58.1 m/s$ is the final velocity of the ball

$u=0$ is the initial velocity of the ball

$S=0.15 m$ is the distance during which the force is applied

$a$ is the average acceleration

Rearranging the formula and substituting the numbers, we find:

$a=\frac{v^2-u^2}{2S}=\frac{(58.1 m/s)^2-0}{2 (0.15 m)}=1.13 \cdot 10^4 m/s^2$

Answer 2

The average acceleration of the ball during her serve : a = 11232.68 m / s

Further explanation  

Regular straight motion is the motion of objects on a straight track that has a fixed speed  

Formula used  

$\large{\boxed{\bold{S=v\times\:t}}}$

S = distance = m  

v = speed = m / s  

t = time = seconds  

Straight motion changes regularly are the straight motion of objects that have a fixed acceleration  

Formula used  

$\large{\boxed{\bold{St=vo.t+\frac{1}{2}at^2}}}$

V = vo + at  

Vt² = vo² + 2aS  

St/S = distance on t  

vo = initial speed  

vt = speed on t  

a = acceleration

The initial speed of the ball when hit = 0

Ball speed = 130 miles / h (209 km / h) = 58.05 m/s

Distance traveled by the ball = 0.15 m then

Vt² = vo² + 2as

58.05² = 0² + 2. a.0.15

3369.8 = 0.3.a

a = 11232.68 m / s

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Keywords:  the average acceleration,fixed acceleration, fixed speed