# Stopping a car.

1. Sep 13, 2006

### physics newb

OK, I have some homework, due soon, and could use some help. If you can give me an answer and show me how you got it, great, but I'm here to learn too.
A car with a mass of 1300kg is initially moving at a speed of 40 km/h when the brakes are applied and the car is brought to a stop in 15 m. Assuming that the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied?
So there are 4 parts to this problem. I'm here to learn, but I'm also crunched for time, so any help would be appreciated.

2. Sep 13, 2006

### stunner5000pt

u have to tell us how you have tried to solve the problem yourself...

3. Sep 13, 2006

### physics newb

I haven't, I don't even know where to begin. That's why I'm here. Once I get started, things should start clicking. Does the mass of the car have anything to do with how long it will take to slow down?

4. Sep 14, 2006

### physics newb

Does anyone have an idea as to what the next step is?

5. Sep 14, 2006

### Tomsk

For a) you need the acceleration, and you only know the initial and final velocities, and the distance, so you need an equation that only uses these variables. Do you know what it would be? Don't forget to multiply a by the mass to get the force. Then for b), you need the time, so since you have the acceleration, you can use another equation of motion, and solve it for time. For c) and d), go back and solve these 2 equations for x and t, with the new initial speed.

6. Sep 14, 2006

### physics newb

$$v^2 = v_0^2 + 2 a \Delta x$$
Would this be it?

7. Sep 14, 2006

### physics newb

Time is not my luxury here. What is the next thing I do?

8. Sep 14, 2006

### Hootenanny

Staff Emeritus
What confuses you? Tomsk's post seems to clear it up pretty well. Your equation is the correct one to use for (a).