# Finding acceleration and distance

1. Sep 21, 2010

### kari82

A soccer ball is released from rest at the top of a grassy incline. After 6.3 seconds, the ball travels 51 meters. One second later, the ball reaches the bottom of the incline.
a) What was the ball's acceleration? (Assume that the acceleration was constant.)
b) How long was the incline?

a)
at rest t=0 and yi=0
at t=6.3 yf=51 m
yf-yi= vi(t)+1/2(a)(t)^2
51m=0(6.3)+1/2(a)(6.3 s)^2
[(51m)(2)]/39.69s^2=a=2.57 m/s^2

b)
a(constant)=2.57m/s^2
t(final)=7.3s

deltay=vi(7.3s)+1/2(a)(t)^2
deltay=0+1/2(2.57m/s^2)(7.3s)^2
deltay=68.5m

2. Sep 21, 2010

### Staff: Mentor

Looks fine to me. Why do you think it doesn't make sense?

3. Sep 21, 2010

### kari82

This is my first physics class (university physics) and I always feel like I'm missing something.. I like to have confirmation from someone that knows what they r doing ;-)

Can you give me a hint on question c please? Thanks!!

4. Sep 22, 2010

### Staff: Mentor

Did you post a c) question? Did you post your attempt at a solution for whatever c) is?

5. Sep 22, 2010

### kari82

Question c was in a different post.. I'm sorry!! I made a mistake..