# Decelerating distance and magnitude of impulse of a car

1. Apr 29, 2014

### Juls808

A 1,000kg car traveling with a velocity of +20.0meters per second decelerates uniformly at -5.00 meters per second per second until it comes to rest.

What is the total distance the car travels as it decelerates to rest?

What is the magnitude of the impulse applied to the car to bring it to rest?

2. Apr 29, 2014

### TheAustrian

What exactly do you require help with? I mean, where in your working did you get stuck?

3. Apr 29, 2014

### Juls808

I don't know which equations to use.

4. Apr 29, 2014

### BiGyElLoWhAt

You have a velocity and an acceleration. How do those relate to displacement? (or distance in this case)

5. Apr 29, 2014

### Juls808

So it would be velocity=distance/time and it would be that simple because its uniform? And impulse=force*time?

It can't be that simple because that gives me a distance of 100 meters.

6. Apr 29, 2014

### TheAustrian

You have to take the (de-)acceleration into account. So you're going to need to relate your acceleration to distance, with having an initial velocity in the opposite direction. (so you could take your acceleration as negative, and your velocity as positive)

7. Apr 29, 2014

### dauto

Why don't you show us your calculations step by step? That way we can pinpoint any mistakes. Just telling us you found a distance of 100 meters (which is not correct) with no explanation whatsoever about how you got to that number is not very enlightening. And please write the equations down before plugging in any values so we know what equation you're using. (By the way, do that on your homework and tests as well. Your professor will greatly appreciate).

8. Apr 29, 2014

### BiGyElLoWhAt

You're losing me. Impulse is a change in momentum, and is also $\int \vec{F}dt$ but I have no idea why you're using that. What in this problem says, "use momentum" to you? (Other than the mass that's given, keep in mind physics profs are tricky ;-] )

Ok so you figured out that $|v| = \frac{\Delta x}{\Delta t}$ which works for this, but really what you want is $\vec{v} = \frac{d\vec{x}}{dt}$ and you have another quantity in there, acceleration $\vec{a}$

Now how does acceleration relate to what we have?