# Homework Help: Uniformly accelerated motion- a weird kangaroo prob

1. Sep 28, 2004

### bjr_jyd15

Uniformly accelerated motion-- a weird kangaroo prob

Ok, I have this simple Question:

A kangaroo can jump straight up 2.5 m -- what is its takeoff speed?
How can I solve that given just one variable. Well, I know that Vi=0, right?

Also, if you anyone can help me try to solve this one?:

A car in an auto accident left skid marks 290 m long. Assuming an average acceleration of -3.9 m/s/s (that is, -0.4 g), calculate the Jag's speed when the brakes locked.

I know acceleration and scalar distance, but I don't know anything about the initial speed and such. So how can I choose a formula?

Thanks

2. Sep 28, 2004

### Pyrrhus

For first problem, review Free Fall.

For the second problem, what happens when you apply the brakes?

3. Sep 28, 2004

### bjr_jyd15

When the brakes applied, the car slows, right? And skids 290 m. I'm not sure where to go from there

4. Sep 28, 2004

### punjabi_monster

This is how i believe the question is solved:
Vi=?
Vf=0m/s
a=-3.9m/s^2
t=X
d=290 m

use thsi formula
Vf^2 = Vi^2 + 2ad

5. Sep 28, 2004

### tyco05

I'm an Aussie, so I should be able to help with the Kangaroo question!

First off, you have more than one variable.

(Assuming that the kangaroo is on the Earth) The acceleration can be given by g. (ie use -9.8 m/s/s)

The displacement is 2.5m, and Vf = 0 (at the top of the jump, the kangaroo's velocity is zero).

Vi is the unknown, that you want to calculate.

I believe the constant acceleration formula you might need is:

$$v_f^2 = v_i^2 + 2as$$

where Vf and Vi are the final and initial velocities respectively.
a is the acceleration
s is the displacement

also don't forget the direction of things. (if you use +2.5 for the displacement, you MUST use -9.8 for the acceleration)

6. Sep 29, 2004

### bjr_jyd15

thanks

Great, thanks Tyco and punjabi. I got the problems, and I actually understand them.