# Homework Help: Cat dropped from a Shelf, deceleration time?

1. Oct 4, 2011

### rexexdesign

1. The problem statement, all variables and given/known data
A cat drops from a shelf 3.7ft above the floor and lands on all four feet. His legs bring him to a stop in a distance of 12cm.

Part A: Calculate his speed when he first touches the floor (ignore air resistance).
Part B: Calculate how long it takes him to stop.
Part C: Calculate his acceleration (assumed constant) while he is stopping, in m/s2.
Part D: Calculate his acceleration (assumed constant) while he is stopping, in g’s.

2. Relevant equations
For Part A: V2y = V2y0+2ay(y-y0)
I am stuck at part B, I know I need to find the time that it takes to slow down, right? This is my first week of Physics and I am not fluent in using the formulas or knowing what Vyo always is and what to use for acceleration. I tried to use this formula: y = yo + vo t + ½ a t2

3. The attempt at a solution
Part A (correct) 0 + (2*32.2ft/s2) * (3.7ft-0) = 4.7m/s (which I entered in my HW and it's correct)
Part B When I use Vo = 4.7m/s and a=9.8m/s2 and V=0 (because the final velocity would be 0 when the cat stopped completely, right?) I get 0.16s and the program said it was wrong. I also got another result by changing the equation of 0.11s which was wrong as well. Finally I looked in my physics book and found the SAME problem, but the cat jumped from 4.0ft instead of 3.7ft. They got 4.9m/s velocity as the first answer, and 0.049s as the time to stop... so I though well it must be 0.047s for me then... wrong. Also didn't work. I only have 3 attempts left on this question and I am stuck. The book got an acceleration of 100m/s2 for part C and 10g for part D, which even confused me more. How can it have such a high acceleration?

Any help would be much appreciated, the HW is due tomorrow and I was up till 1 AM last night trying to figure this out, no problem in the book deals with a deceleration within a certain distance like this.
I am sure this is easy to most of you, but I am just getting my feet wet with Physics!

Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 4, 2011

### Staff: Mentor

The cat does not decelerate at g. That's his falling rate of acceleration. He's got to stop in a much shorter distance than he fell from the shelf.

Can you think of another kinematic equation that involves initial speed, final speed, acceleration and distance?

3. Oct 4, 2011

### rexexdesign

Thank you gneill for your quick reply. So you are saying a≠g so then I would have two unknowns, the time and the acceleration?
Part C is asking for the acceleration, but I need to get the time first, correct?

I have 4 constant acceleration equations from class, none of which have an equation with time but without acceleration. What am I missing?

I know you are not wanting to just give me the answer and I am not expecting that, as I want to learn it myself, I think I am just too much of a noob to get onto the right track, your help is appreciated.

4. Oct 4, 2011

### Staff: Mentor

You have (at least) two choices. You can calculate the acceleration first and then apply the kinematic equation d = vo*t - (1/2)*a*t2 to find the time, or you can find the time first.

To find the acceleration first you can use a common kinematic equation that relates initial and final velocities with the acceleration and distance.

To find the time first you might want to recall what the average velocity is when uniform acceleration is involved.

5. Oct 4, 2011

### rexexdesign

Thank you gneill, I will take a closer look. I guess my mistake was that I used an equation with 2 unknowns, I think I should be able to figure this out. This forum is awesome and I hope to be able to return some favors once I learn Physics in the next 3 quarters :D

6. Oct 4, 2011

### rexexdesign

I got it, thanks so much for the pointers. I actually didn't like the equations I had, so I googled some transformed equations and this one was really easy:
d=vi+vf/2 *t
I know my initial velocity is 4.7m/s and the final velocity must be 0 to be stopped. Then I have the distance of 0.12m. That's how I got time. (0.051s) (Part B)

There was a little typo in your equation vo*t - (1/2)*a*t2 (it's actually + not -, right?)

Anyway, when I used. d=vi*t + 1/2*a*t2
I got that a was 92m/s2. (Part C)

Finally I calculated g's by dividing 92m/s2 by 9.8m/s2 = 9.4g's

Thanks so much for your help. I will make sure to write down the equations I had from class in the new format, they are a bit easier to understand. Using Vf instead of V and Vi instead of Vo is a bit easier imo when one gets started with this. Hope this threat helps some other students as well. I bled some points on my homework, but certainly learned a lot!

7. Oct 4, 2011

### Staff: Mentor

I'm glad it all worked out.

You'll see various notations for equations in different materials. The trick is to recognize the form of the equation to see through the camouflage of different variable names. This comes with repeated use.

Regarding the equation that I wrote, I assumed that the acceleration was specified as a magnitude (unsigned), so the sign was incorporated into the equation itself. The important thing is to recognize the form of the equation required and apply it accordingly to the problem at hand. That's more than half the battle.

Keep up the good work!

8. Jul 15, 2012

### DanielHen

He touches the floor at 3.7 ft - 12 cm, so shouldn't it be (2*32.2ft/s2) * (3.7ft-0.4ft)?

9. Jul 15, 2012

### azizlwl

The floor is 3.7ft. below. Less than that is empty space.

10. Jul 15, 2012

### DanielHen

Of course. Thanks!

11. Sep 27, 2012

### kcaz

I have been trying to figure out part (b) to that question with the cat except in my course the numbers were slightly different. ...

A cat drops from a shelf 4.4ft above the floor and lands on all four feet. His legs bring him to a stop in a distance of 12cm .....
Calculate how long it takes him to stop....

I cannot figure out how to solve this one? Can someone please lend a hand?