# Homework Help: Deceleration of a car.

1. Dec 8, 2011

### Kujanator

1. The problem statement, all variables and given/known data
The braking distance is the distance the car travels while decelerating once the brakes have been applied. A car of mass 450kg is travelling at a speed of 70MPH (31.3ms-1) when the driver makes an emergency stop.
Calculate the deceleration of the car (Assume a uniform deceleration)

We're also told that at 70mph/31.3ms-1 that the breaking distance is 75 m.

3. The attempt at a solution
I did initially try to do this using the equations of motion but that gave me a huge answer, and didn't take into account the mass of the car so I didn't think it was right.

Any help is hugely appreciated!

2. Dec 8, 2011

### Staff: Mentor

You should show your attempt, even if it has an error or two. That way we will know how to help.

3. Dec 8, 2011

### Kujanator

Ok, this is what I did.

S=75
U=31.3
V=0
A=?
T=?

s=u²+2as
0=31.3²+2*a*75
0=979.69+150a
-979.69/150=a
a=-6.53 ms-2

Which looking back doesn't look too bad, when I did it first time I think I accidentally must have typed the wrong numbers into my calculator because I got -1000ms-2. I'm just thinking I did it wrong because I didn't use the mass.

4. Dec 8, 2011

### Staff: Mentor

Looks good.

One small nit though, in your first equation you put "s" on the left where you meant to put "V2". Clearly a typo, since you correctly substituted the value of V2 in your next line. So overall, well done.

5. Dec 8, 2011

### Kujanator

There's just one other part:
The car is then filled with passengers so that its mass is doubled. Calculate its stopping distance when travelling at 70MPH. Assume the force calculated in b(ii) remains constant.

I calculated the force as:
F=ma
F=450*6.53= 2938.5N

How do I go about doing this is it:

2938.5 = 900 * a
2938.5/900 = 3.625

So:
v²=u²+2as
0=31.3²+2*3.625*s
-979.69/7.25=s
s=135 metres.

Thanks again!

6. Dec 8, 2011

### Staff: Mentor

Check your value for the new acceleration. Looks like a finger problem. Otherwise you're method works okay.

You could skip the force calculation if you realize that since F = MA, then A = F/M. Therefore if you double M you must halve A.