Uniformly accelerated motion-- a weird kangaroo prob

[bjr_jyd15]

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?

Please Help.

Thanks

[Cyclovenom]

For first problem, review Free Fall.

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

[bjr_jyd15]

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

[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

[tyco05]

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

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)

[bjr_jyd15]

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