# Homework Help: Car Accelerating

1. Sep 2, 2014

### DustyGeneral

Hello all, here is my problem:

A car accelerates, starting from rest, with a speed that is given by:

v(t)=vm(1-e-at)

a) What is the top speed of the car? Explain why.

b) How far does the car travel in time t?

c) Suppose the car can accelerate from 0 to 60 mph in 2.9s, and has a top speed of 195 mph. Imagine a mile-long race between two of these cars, with the same finish line, but with a different starting line: One drives along the ground towards the starting line at a point one-mile away, while the other is dropped out of a plane one mile above the ground. Which one reaches the finish line first? (ignore any air resistance for the falling car).

That is the entire problem. I just need help getting started.

First, what does the vm represent in the speed equation?
Second, how exactly do I calculate the top speed?

I also suppose that I will need the position and acceleration equations which I derived from speed equation for b and c respectively.

I got:

x(t)=vm*((e-at)/a+t)
v(t)=vm*(1-e-at)
a(t)=vm*(ae-at)

2. Sep 2, 2014

### Staff: Mentor

Think of it as a constant. Once you answer part a, the notation will make more sense.

You won't be able to get a numerical value, if that's what's throwing you off. Answer in terms of the constants given.

You are given the speed as a function of time. If you graphed it, what would that function look like? What's its maximum value?

3. Sep 2, 2014

### DustyGeneral

I know I won't get a numerical value as the top speed. I just do not know what path to take to manipulate v(t) to get top speed.

4. Sep 2, 2014

### Staff: Mentor

Does the speed increase or decrease as time increases?

What's the maximum value of 1-e-at ?

5. Sep 2, 2014

### DustyGeneral

Increase in time=Increase in speed.

Max value of 1-e-at = 0

6. Sep 2, 2014

### Staff: Mentor

Good!

Really? So the car doesn't even move? (Think that one over. )

7. Sep 2, 2014

### CWatters

Are you sure?

8. Sep 2, 2014

### DustyGeneral

Nevermind. Duh. I was thinking something completely different. So what is the a in the exponent? Acceleration due to gravity? If so then the max value would be 1, but that doesn't make sense.

9. Sep 2, 2014

### Staff: Mentor

Treat 'a' as just another constant.

Makes sense to me. At what time will the car have max speed? What is that max speed?

10. Sep 2, 2014

### DustyGeneral

So what I'm getting is that:

v(t)=vm(1-e-at)

We've said that the max value for (1-e-at) = 1

So that means at max v(t)=vm

In order to get that t=∞.

11. Sep 2, 2014

### Staff: Mentor

Perfect! As time goes on, the speed gets closer to the maximum value of vm.

Perhaps now you can guess what the 'm' stands for.

12. Sep 2, 2014

### DustyGeneral

I figured that's what it stood for I seem to have a habit of jumping to conclusions like that early on in the problem and confuse my self because I get determined to make that scenario be the case (right or wrong.)

So part a is understood. Part b would be the distance equation setting t=∞?

13. Sep 2, 2014

### Staff: Mentor

They don't want the distance at infinite time, but at time "t". (You've already solved that one in your first post.)

14. Sep 2, 2014

### DustyGeneral

See what I mean? Get ahead of myself and flustered and try to make a certain scenario fit. Then for part c I am given acceleration, the time for that acceleration, and top speed. Plug and chug then. Very helpful Al.

15. Sep 2, 2014

### Staff: Mentor

Correction: You've almost solved it. Don't forget about the constant of integration.