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worthbeads · Feb 16, 2007, 05:04 PM

This is just a problem that has been bugging me. This is not me homework.:D It was on a car commercial.

A car company claims they can drop a car 4000 feet and have another car on the ground. I'm pretty sure the car on the ground traveled the same distance. Assuming the car being dropped accelerates at 10 meters per second per second, and the car on the ground is already traveling at a constant speed, can the car on the ground travel faster? How would you work this problem? Is there enough information?

It's just bugging the heck out of me!:confused:

Capuchin · Feb 17, 2007, 02:05 AM

I don't really know what the terminal velocity of a car would be, which is quite internal to the problem :/

worthbeads · Feb 17, 2007, 08:43 PM

How about we assume there is no limit on how fast you can go. Now how would you solve it?

Capuchin · Feb 18, 2007, 03:28 AM

okay so

$v^2 = u^2 + 2as$

$v^2 = 0 + 2*32.15*4000$

v = 507 ft/s

that's 154 m/s

Average speed = 154/2 = 77m/s = 0.077km/s = 277km/hour

The car on the ground would have to be going a constant speed of 277km/h or 172 mph

Doable I think!

It would be slower than this due to air resistance. It seems very doable and not really a stunning performance :)

worthbeads · Feb 18, 2007, 08:37 AM

Originally Posted by Capuchin

okay so

$v^2 = u^2 + 2as$

$v^2 = 0 + 2*32.15*4000$

So what do the variables $a$ and $s$ and $u$ stand for, and why is the equation $v^2 = u^2 + 2as$ ?

Capuchin · Feb 18, 2007, 08:53 AM

That's a classic equation of motion for uniform acceleration.

(put "suvat" into google)

a is acceleration, s distance, u initial speed.

worthbeads · Feb 19, 2007, 06:02 PM

So how do you get 32.15 as acceleration?

Capuchin · Feb 19, 2007, 11:21 PM

That's 9.8m/s/s in ft/s/s (a standard result)