Acceleration again

[Omid]

Two cars are in a drag race. The first starts and continues to uniformly accelerate up to the finish line some time later. The second car follows the first out of the starting box 1s afterwards traveling exactly as it did. Will the cars be the same distance apart throughout the race? How much time will there be between their successive crossing of the finish line ?

I think the answer to the first question is yes, and to the second one : 1s.
What do you think about that ?
Thanks

[faust9]

If this where changed a little to say "Two balls are dropped from the same height but with a 1 second delay between the release of each ball" Would the distance between the balls remain constant?

[Omid]

Yes it would remain constatn.

[faust9]

No... What is the definition of acceleration?

[Tide]

Are you sure?

[faust9]

Are you sure?

Who are you asking?

[BobG]

Two cars are in a drag race. The first starts and continues to uniformly accelerate up to the finish line some time later. The second car follows the first out of the starting box 1s afterwards traveling exactly as it did. Will the cars be the same distance apart throughout the race? How much time will there be between their successive crossing of the finish line ?

I think the answer to the first question is yes, and to the second one : 1s.
What do you think about that ?
Thanks
The answer to the first is 'No'. The answer to the second is '1 sec'.

If the cars are accelerating at a constant rate, they cover more ground each second than they did the previous second.

If they are accelerating at 2 m/s^2:

After 1 sec, the first has moved 1 meter - the second 0.
After 2 sec, the first has moved 4 meters - the second 1 meter.
After 3 sec, the first has moved 9 meters - the second 4 meters.
And so on.

[faust9]

\vec{a}=\frac{dv}{dt}

At any given time the velocities of the two falling bodies will not be the same. The one dropped first will have a greater V at time 1 than the second body. If you look at acc. of each body with respect to the earth the acc. will be the same. Look at the acc. of one body with respet to the other and you see that the acc is constantly changing.

\vec{a}_{rel}=\frac{dv_{rel}}{dt}

When doing comparisons like this you need to look at the bodies in relation to each other and not in relation to a third body (usually).

Good luck.

[Omid]

Thanks a lot

[Omid]

So they are closest to each other after first second, aren't they ?

[Tide]

Who are you asking?

Sorry - I was asking Omid who stated the separation would be constant. I didn't notice you had slipped in there before I posted it.