# Time at given speed in freefall

1. Jan 24, 2004

### KingNothing

Here is a problem from my book:

If a car rolls gently (initial velocity = 0) off a vertical cliff, how long does it take to reach 90km/h?

It seems easy, and I have been breezing through all teh other ones like it's my job, but for some reason this one gets me. I knwo it's possible, but I jsut don't know how. The answer is supposed to be 2.6 seconds, but I somehow always wind up dividing 90 by 9.8. Any help?

2. Jan 24, 2004

### chroot

Staff Emeritus
Are you familiar with the expression:

$$v(t) = v_0 + \frac{1}{2} a t^2$$?

Just solve it for t.

- Warren

3. Jan 24, 2004

### KingNothing

Alright, thanks...but I still seem to get the wrong answer consistently. I do:

90=0.5(9.8)t^2
90=4.9t^2
18.37=t^2
t=4.29

what is going wrong?

4. Jan 24, 2004

### chroot

Staff Emeritus
Convert 90 km/hr into m/s first, so that your units will agree.

- Warren

5. Jan 24, 2004

### KingNothing

Thanks, I don't know how I skipped that. That's what happened on the quiz, too. Did all of them perfectly, then on one forgot to convert.

Also, is it jsut me, or should I really end up with about 2.26 seconds?

Last edited: Jan 24, 2004
6. Jan 25, 2004

### Julian Solos

I get 2.258769757 on my calculator using g = 9.8 m/s^2.

Does the problem explicitly say we can ignore air resistance?

7. Jan 25, 2004

### HallsofIvy

Chroot: I must confess that during the several days I spent in calculus (and physics) class, I never saw an equation that said
"$$v(t) = v_0 + \frac{1}{2} a t^2$$"

I do recall seeing one that said "$$v(t)= v_0+ at$$" and one that said "$$d(t)= d_0+ v_0t+ \frac{1}{2}at^2$$".

Assuming no air resistance and that a= -9.8 m/s2, with v0= 0, v(t)= -9.8 t m/s. Since there are 1000m in a km and 60*60= 3600 sec in and hour, -90 km/hr= -90 *1000/3600= -25 m/s.
-25= -9.8 t so t= 25/9.8= 2.55 seconds.

Looks pretty easy to me (as long as you don't listen to Chroot!).

(C'mon, who is it that's using Chroot's good name?)

8. Jan 25, 2004

### Jimmy

I'm sure it was just an oversight. .

Last edited: Jan 25, 2004
9. Jan 25, 2004

### chroot

Staff Emeritus
GAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAH!

*slinks into a corner and hides*

- Warren

10. Jan 25, 2004

### PrudensOptimus

Or you can use

v^2 = v0^2 + 2a&Delta;x.

Solve for &Delta;x and then time is just &Delta;x/v.

11. Jan 25, 2004

### Warr

Optimus, you can't do that because once you've solved for the distance its fallen, dividing by v means you are assuming that the velocity is constant, which it obviously isn't.

12. Jan 27, 2004

### Julian Solos

GAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAH!

I blindly assumed chroot's equation was correct.

Give me back the two minutes of my life I wasted!

13. Jan 27, 2004

### Julian Solos

Besides, x is not given.

14. Jan 27, 2004

### Jimmy

Actually, under a constant acceleration, you can compute the time using the average velocity.

Given the final velocity and displacement, to solve for t:

$$t = \frac{2 d}{V_f}$$ or

$$\frac{V_f}{2} = \frac{d}{t}$$

Vf/2 when Vi=0 and acceleration is constant gives the average velocity.

Last edited: Jan 27, 2004
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