Evan,
I didnt intend to start a ping pong match !
You could time your car to a few different speeds.
Then draw a graph of speed (y axis) v/s time (x axis)
Then calculate the area under the graph up to the speed you are interested in. That gives you the distance traveled to get to that speed.
Sorry to be a PITA but thats the correct way to do it.

The previous comment about the shape of the graph is about right.
it would be of a general form Y= kX^(1/2)
Using BC's converted amounts ~ that is time (X) in sec & your speed in ft/sec, you can work out K.
Then integrating for zero to a given time (20 in your case?)

S=(2/3)xkx(20)^(3/2)

That should give you an answer close to the actual value.

Way too sensible.

Leathermang and BC are correct.


...except that speed is different than velocity.

Only if acceleration is constant.Only relevant if the car does not travel in a straight line.

Actually there is NO right or wrong answer. If you assume infinite acceleration or infinite braking power then you can pick a number, any number, will be the right answer. In real life, it would be ..., I give up.

I suppose every one flunk their test.

all correct, but we were given simple parameters and are left to assume kinematics with 2-3 significant figures.

I didn't read the whole thread... sorry if you already hashed this out.

That's not true. The right answer is that he provided insufficient information to make the calculation. The wrong answer is that acceleration has nothing to do with it. With a bit of research about the acceleration characteristics of similar vehicles, one could write an equation that would give a pretty good estimate. To use the average velocity probably underestimates the distance traveled because I suspect that the acceleration is greater at the beginning of the trip than it is at the end.

Okay, now we've done it . . . we've made math a hot topic! Well done gentlemen, well done!

"pretty good estimate" = not a precise answer
"Insufficient info" = use your imagination
"acceleration has nothing to do with it" = assumption, assume to be constant

If you ask a grade school student, he/she will come up with 1466.67' as they use the most simple logic to get to their answer. They do not dwell into an ivory tower like most members here. There are better things in life to do then trying to fit an answer to a question.:D

would have been faster just to drive it.......

And the only truly accurate way. The figure would vary from car to car - there is no perfect equation for it.

Rocket science is indeed rocket science but at least they have one accelerating force that is known to be constant: gravity. During liftoff, air resistance and uneven acceleration from thrust must be contended with but I'm guessing that the thrust from rockets is more constant. No gear changes.

When those booster rockets quit and drop off... I would call that a ' gear change '.....LOL

Here ya go:

D = 1/2 a t2 + ViT where D = distance, a = acceleration, Vi= initial velocity and t= time


D= 1/2 32 ft/sec 2 (20sec) 2 + 0 (20)

D = 16 ft/sec2 (400sec 2 ) = 6400 feet

so, you dropped the car off a cliff and it takes 1.21 miles to get reach 100mph?

whats terminal velocity for this vehicle, or CD?

How many cattle can we fit in a boxcar?

First we assume a square cow.