# Sports Car Acceleration Question

1. Jan 14, 2010

1. The problem statement, all variables and given/known data
A sports car moving at constant speed travels 110m in 5.0s. If it then brakes and comes to a stop in 4.0s. what is its acceleration in m/s^2? Express the answer in terms of "g's." where 1.00 g = 9.80 m/s^2.

2. Relevant equations
I thought about using the equation a=v/t

3. The attempt at a solution
So I tried it sort of I am just sort of lost I am not really sure how to start....

2. Jan 14, 2010

### rock.freak667

You need to use the kinematic equations

v=u+at

v2=u2+2as

s=ut+1/2at2

Which one do you think you will have to use?

3. Jan 14, 2010

What does the u stand for, that's not in my book?

4. Jan 14, 2010

### Gear300

The car initially had a velocity of 110m/5s (22m/s) and came to a stop (0m/s) in 4.0s. The acceleration is the change in velocity over a time interval.

5. Jan 14, 2010

So would we use the v=u+at since I have the 4.0s and the 22m/s? but what would go in for the u?

6. Jan 14, 2010

### Gear300

u is initial velocity and v is final velocity.

7. Jan 14, 2010

Okay I figured it out thanks so much the answer was -5.5 m/s^2 and then in g's it was -0.56 g's

8. Jan 14, 2010

### jasper10

take u (initial speed) = 110 / 5 = 22m/s
v = 0
t = 4

v = u + at
0 = 22 + 4a
4a = -22
a = -5.5m/s^2
a = (-5.5/9.8)g = -0.56g