Time at given speed in freefall

[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 just 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?

 

[chroot]

Are you familiar with the expression:

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

Just solve it for t.

- Warren

 

[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?

 

[chroot]

Originally posted by Decker
what is going wrong?

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

- Warren

 

[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 just me, or should I really end up with about 2.26 seconds?

 

[Julian Solos]

Originally posted by Decker
Also, is it just me, or should I really end up with about 2.26 seconds?

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

Does the problem explicitly say we can ignore air resistance?

 

[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?)

 

[Jimmy]

I'm sure it was just an oversight. .

 

[chroot]

GAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAH!

*slinks into a corner and hides*

- Warren

 

[PrudensOptimus]

Or you can use

v^2 = v₀^2 + 2aΔx.

Solve for Δx and then time is just Δx/v.

 

[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.

 

[Julian Solos]

Originally posted by chroot
GAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAH!

*slinks into a corner and hides*

- Warren

GAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAH!


I blindly assumed chroot's equation was correct.

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

 

[Julian Solos]

Originally posted by 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.

Besides, x is not given.

 

[Jimmy]

Originally posted by 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.

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.