Car Acceleration: Top Speed, Travel Distance & Mile-Long Race

[DustyGeneral]

Hello all, here is my problem:

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

v(t)=v_m(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 v_m 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)=v_m*((e^-at)/a+t)
v(t)=v_m*(1-e^-at)
a(t)=v_m*(ae^-at)

 

[Doc Al]

First, what does the v_m represent in the speed equation?

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

Second, how exactly do I calculate the top speed?

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?

 

[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.

 

[Doc Al]

I just do not know what path to take to manipulate v(t) to get top speed.

Does the speed increase or decrease as time increases?

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

 

[DustyGeneral]

Increase in time=Increase in speed.

Max value of 1-e^-at = 0

 

[Doc Al]

Increase in time=Increase in speed.

Good!

Max value of 1-e^-at = 0

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

 

[CWatters]

Increase in time=Increase in speed.

Max value of 1-e^-at = 0

Are you sure?

 

[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.

 

[Doc Al]

So what is the a in the exponent? Acceleration due to gravity?

Treat 'a' as just another constant.

If so then the max value would be 1, but that doesn't make sense.

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

 

[DustyGeneral]

So what I'm getting is that:

v(t)=v_m(1-e^-at)

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

So that means at max v(t)=v_m

In order to get that t=∞.

 

[Doc Al]

So what I'm getting is that:

v(t)=v_m(1-e^-at)

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

So that means at max v(t)=v_m

In order to get that t=∞.

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

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

 

[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=∞?

 

[Doc Al]

Part b would be the distance equation setting t=∞?

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

 

[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.

 

[Doc Al]

(You've already solved that one in your first post.)

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