Question

Starting from rest at home plate, a baseball player runs to first base 90 ft away. He uniformly accelerates over the first 10 ftto his maximum speed, which is then maintained until he crosses first base. If the overall run is completed in 4 seconds, determine his maximum speed and the time duration of the acceleration.

Answer 1

Maximum speed = 25 ft/s

Time during acceleration = 0.8 s

Explanation:

Let max speed =
`v_m`, time during acceleration =
`t`

`\text{distance} = \text{average speed}*\text{time}`

`\text{average speed}=\frac{\text{initial speed}+\text{final speed}}{2}`

During the acceleration, initial speed is 0 (since he starts from rest) and the final speed is the max speed. He travels 10 ft during this period. So

`(v_m+0)/(2)* t =10`

`v_mt=20`

For the rest of the travel, he spent
`4-t` seconds travelling at
`v_m`. This distance is
`90-10=80` ft.

`v_m*(4-t)=80`

`4v_m - v_mt =80`

But
`v_mt = 20` from previously. Hence,

`4v_m-20=80`

`4v_m=100`

`v_m=25`

Substitute for
`v_m` in
`v_mt=20`

`25t=20`

`t=0.8`

Therefore, the maximum velocity is 25 ft/s and the time during acceleration is 0.8 s.