# Am I setting up this kinematic equation properly?

A race car starts from rest and goes a quarter mile (1/4 mi) in 12.1 seconds. Assume the acceleration of the car is constant. (a.) What is the acceleration of the car? (b.) What is the final speed of the car? (c.) If you hit the brakes, how far would it take to stop the car if the breaks cause an acceleration of -2.00 m/s2?

I believe I have parts a and b right, however I am not confident with my process of solving part C.

1/4 mi ≈ 402.3 m

a.) acceleration = 5.5m/s^2
b.) Final speed (at 12.1 seconds) = 66.5 m/s

c.) For part C, I reset the known values to:
a = -2.00m/s2
V0 = 66.5m/s
V = 0 (because solving for car stop which is zero velocity)
X0 = 402.3m

The equation I used: v2 - V20 = 2a(x - x0)
The answer I got was 1507.8m = x
For a car going 66.5m/s that is slowing down at -2.00 m/s2 does this answer seem practical? I am struggling to check my work here.

Many thanks.

## Answers and Replies

haruspex
Science Advisor
Homework Helper
Gold Member
X0 = 402.3m
It asks how far the car goes after hitting the brakes. The distance already travelled is not relevant.

It asks how far the car goes after hitting the brakes. The distance already travelled is not relevant.
So if I change X0 to be = 0 the new answer I get is 1105.5m. This still seems rather far for a car to stop from 66.5m/s to 0m/s. Is -2.00m/s^2 simply just a very slow breaking acceleration?

haruspex
Science Advisor
Homework Helper
Gold Member
Is -2.00m/s^2 simply just a very slow breaking acceleration?
It is rather modest. Safe but serious braking is about 0.5g on a dry road. Emergency braking could go as high as 0.7-1g, depending on the vehicle.

And it is "braking", using "brakes", not "breaking".

• Orodruin