Constant acceleration of baseball problem

[iPhotonHQ]

Homework Statement

"The fastest measured pitched baseball left the pitcher's hand at the speed of 45 m/s. If the pitcher was in contact with the ball over a distance of 1.50m and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?"

- I am doing this problem as part of my studying and I don't think that the textbook gives an answer to it. I am wondering if I did it correctly... Thanks!

- Assume one-dimensional motion.

Homework Equations

x is displacement (because my professor likes to use x for that).
x = ((v_i + v_f) / 2)t
a = (v_f - v_i) / t

The Attempt at a Solution

v_f = 45 m/s
v₀ = 0 m/s
x = 1.5m
x = (45 + 0)/2 *t
x = 22.5t
1.5 / 22.5 = t
t ≈ .067

a = 45m/s / .067s
a ≈ 671.642 m/s²

 

[garylau]

https://www.google.com.au/search?q=...ed+pitched+baseball+left+the+pitcher's+hand+"


i found there are lot of similar question here~

 

[garylau]

https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.1040713.html
It starts at zero and accelerates to 45 m/sec --> avg speed = 22.5 m/sec
-----

t = d/r = 1.5/22.5 = 1/15 second
========
Acceleration = 45 m/sec/(1/15 sec) = 675 m/sec/sec

 

[iPhotonHQ]

https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.1040713.html
It starts at zero and accelerates to 45 m/sec --> avg speed = 22.5 m/sec
-----

t = d/r = 1.5/22.5 = 1/15 second
========
Acceleration = 45 m/sec/(1/15 sec) = 675 m/sec/sec

Thanks that confirmed my answer!