# Confusion with how to calculate acceleration in this problem

1. Mar 4, 2016

### Frankenstein19

1. The problem statement, all variables and given/known data
What average net force is required to bring a 1500kg car to rest from a speed of 27.8m/s within a distance of 55m?

2. Relevant equations
vf-vo/(time elapsed)

3. The attempt at a solution

I know I need to calculate the acceleration first, so immediately I thought of taking the distance and dividing it by the speed in order to get time, doing that I got 1.92s.

Then when I put the velocity over that time to get the acceleration, I thought i would get the correct acceleration but no. I got -14.48 and the book says its -7. The book uses the formula v^2=vo^2+2ad

Now, my question is: Why doesn't t work my way. Please don't tell me, it's much easier if you use the v^2 formula because I still won't understand since my question is why doesn't the way I thought work.

2. Mar 4, 2016

### BvU

Ah ! but the speed is gradually decreasing all the way to zero !

3. Mar 4, 2016

### Frankenstein19

So I can't just do vf-vi/(v/s)? I'd get 0-55m/(28.7m/s)

4. Mar 4, 2016

### Biker

When you divide distance by time, You are getting the average velocity. it is the average distance that this person cuts over time. Also that time that you got through that method is just the time that this average (constant) velocity will take to pass this distance. In another way, that time assume that the velocity over this distance is constant (doesn't change). If there is no acceleration then you can use this method.

However as BvU mentioned it is decreasing. You are not cutting the same distances over time. So as a result you cant get time through d/v.

Can you explain your equation above? (Vf-vi)/v/s?

5. Mar 4, 2016

### haruspex

Others have pointed out the error in your calculation. I would just like to add a couple of points.
The question should have specified constant acceleration, and therefore a constant force, making the qualifier 'average' redundant.
Average force is defined as $\frac{\Delta p}{\Delta t}=\frac{\int F.dt}{\int .dt}$, where p is momentum. This is consistent with average acceleration, $\frac{\Delta v}{\Delta t}=\frac{\int a.dt}{\int .dt}$.
You need the acceleration to be constant in order to be able to apply $v^2=v_o^2+2ad$.
With non-constant acceleration, we could slow the car to 1m/s in 0.5s (-53.6m/s2), then bring it to rest in 95.6s at a uniform acceleration of -0.01m/s2.
The average acceleration would be -27.8/(.5+95.6)=-0.29m/s2, much less than for constant acceleration.
That's one way, but it's not the only way. You could arrive at the same answer by considering energy. Again, this is only valid for constant force.