# A Grade 11 Kinematics Problem

1. Oct 7, 2007

### aksarb

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

A skier is going 8.2m/s when she falls and starts sliding down the ski run. After 3.0s, her velocity has changed to 3.1m/s. How many seconds after she originally fell did she finally come to a stop? (Assume constant acceleration)

2. Relevant equations

Δd = 1(v1+v2)Δt
2
2aΔd = v2^2 - v1^2

Δd = v1 Δt + 1/2 a (Δt)2

Δd = v2 Δt - 1/2 a (Δt)2

3. The attempt at a solution

Since the question gave me 3 pieces of information, I just tired to sub it into the equations no problem. I realized afterwards that I was only going to find displacement, not the time. But, I already have time, so I have no clue how to get this question. Please help me.

2. Oct 7, 2007

### Kurdt

Staff Emeritus
From the information given you can find the acceleration. Once you have found the acceleration you know the initial speed (8.2 m/s) and the final speed (0 m/s) and you can then solve for time.

3. Oct 7, 2007

### aksarb

So, I would have to use basically get acceleration from v2-v1 then from there use
t
t = v2-v1 to get the answer?
a

4. Oct 7, 2007

### Kurdt

Staff Emeritus
Sorry I missed that equation. Yes you can simply solve from the following equation knowing that the final velocity will be 0.

$$d = \frac{v_i+v_f}{2} t$$

I assumed since the question gave extra information you'd have to work out acceleration.

5. Oct 7, 2007

### aksarb

Im not sure about the d. I do not have a displacement and I am not trying to find it, so how can I eliminate the d so the time can be by itself?

6. Oct 7, 2007

### Kurdt

Staff Emeritus
OK got confused with another thread I'm helping with.

$$v = v_0 + a t$$

You will need to solve for acceleration and then again for time.

7. Oct 7, 2007

### aksarb

Ok, this question is asking for time right? So I just have to find acceleration from the information given then find time of the skier finally comming to a full stop after she originally fell?

8. Oct 7, 2007

### Kurdt

Staff Emeritus
Yes that should do it. I apologise for the confusion. I'm not on form tonight. I've already confused the other person with a kinematics question.

9. Oct 7, 2007

### aksarb

Its ok, thanks for your help, I can finally relax :)

10. Oct 7, 2007

### aksarb

The question has the following data "After 3.0s, her velocity has changed to 3.1m/s." How come I didn't have to use the 3.0 seconds or the 3.1 m/s to get the answer?

11. Oct 7, 2007

### Kurdt

Staff Emeritus
You will need to use that to find acceleration. v0 = 8.2m/s, v = 3.1 and t = 3.

12. Oct 7, 2007

### aksarb

Ok thanks. Can you check the answer for this question? I did

a=v2-v1/t
= 3.1m/s + 8.2m/s / 3.0s
= 3.8 m/s^2

t = v2-v1 / a
= 3.1m/s + 8.2 m/s / 3.8m/s^2
= 3.0s

13. Oct 7, 2007

### Kurdt

Staff Emeritus
be careful with signs.

Remember that when the skier comes to a rest the speed will be 0 m/s. Signs again as well.

14. Oct 7, 2007

### aksarb

Should it be positive because the 8.2m/s is going down due to the regualr sign conventions as left or down being negative? I conserded the sign conventions so the speed became -8.2m/s. Since there is the negative sign, it turned into a positive.

15. Oct 7, 2007

### Kurdt

Staff Emeritus
If you take 8.2 away from 3.1 you will end up with a negative value for acceleration which makes sense since the skier is slowing down. When you solve for time the negative signs cancel so it all works out fine.

16. Jan 20, 2012

First you use V=V(initial)+at to find the acceleration. This will give you -1.71.

Then you plug that into the same equation ^ but this time you make the final velocity equal zero and just find the time.
It works out to 4.8s.

17. Jan 20, 2012