# Linear Momemtum Need dire assistance

1. Nov 1, 2005

### Vivek

This is from my class work and i was not there when the teacher did this.I want to understand this.Please assist

A car with a linear momentum of 3.0 x 10 raised to 4 KG x M/S is brought to a stop in 5.0 sec.What is the manitude of the average braking force?

I really don't get this one.Any help is more than appreciated.

2. Nov 1, 2005

### FredGarvin

Hint: Newton's Second Law can be thought of in this way:
$$\Sigma F = \frac{d}{dt} (mv)$$

3. Nov 1, 2005

### Vivek

The second law is that the acceleration of an object is dependent on two factors - the net force acting upon the object and the mass of the object...

But in relation to the question there are a few variables missing..

4. Nov 1, 2005

### µ³

integral of force-time graph is change in momentum

5. Nov 2, 2005

### HallsofIvy

Staff Emeritus
Most simply, force equals mass times acceleration. Although you are given the initial momentum, rather than the initial speed, since momentum equal mass times speed, and mass stays contant in this problem, just divide the "change in momentum" (from 3.0 x 104 to 0) by the time required.

6. Nov 2, 2005

### FredGarvin

Look at it like this...Starting with

$$F = m a$$ where $$a = \frac{d}{dt} v$$

$$F = m \frac{d}{dt} v$$ since mass is constant, we can pull it inside the derivative

$$F = \frac{d}{dt} (mv)$$ or $$F = \frac{d}{dt} P$$ where P = mv is the linear momentum.

So if the force is equal to the time rate of change of the momentum, how does that fit into your problem?