# Homework Help: Two Kinematics Problems

1. Jan 6, 2008

### vertciel

[SOLVED] Two Kinematics Problems

Hello everyone,

I am having some trouble with the following two kinematic problems and I would appreciate any help or hints.

Thank you!

---

1. Two rugby players are running towards each other. They are 37 m apart. If one is accelerating from rest at $$0.5 m/s^2$$ and the other was already moving at 3.1 m/s and maintains her speed,

a) how long before they crunch together?
b) how fast was the accelerating player going?

My Work :

I am unsure of this, but I think I would need to set two equations equal to each other: each equation representative of each of the two rugby players. However, I do not know which equations I would need to set equal to each other.

2. A police car stopped at a set of lights has a speeder pass it at 27.78 m/s. If the police car can accelerate at $$3.6 m/s^2$$,

a) how long does it take to catch the speeder?

My Work :

I think that the concept is the same; I would need to set one equation for the police car and one for the speeder equal to each other.

Last edited: Jan 6, 2008
2. Jan 6, 2008

### foxjwill

You are correct. You would use a system. Try applying $$v^2 = {v_0}^2 +2a\Delta x$$ to both players.

3. Jan 6, 2008

### Kurdt

Staff Emeritus
The sum of the distance the two travel has to equal 37m for the first one. If you need the time then then the best equation to use is $s= ut +1/2at^2$.

Again for 2.), once the speeder crosses the line and the police car sets off they both have to travel the same distance for the police car to have caught up.

4. Jan 6, 2008

### vertciel

Thank you for your replies.

@Foxjwill, so are you recommending me to use $$(v_2)^2 = (v_1)^2 + 2ad$$?

@Kurdt, are you recommending me to use $$displacement = (v_1)t + \frac{at^2}{t}$$ ? Also, I tried to represent the sum of the distance of the two as such but I did not get the right answer:

$$(v_1)t + \frac{at^2}{t} + v_1t + \frac{at^2}{t} = 37$$

Meanwhile, I will continue to work on the second problem.

Thanks again.

5. Jan 6, 2008

### Kurdt

Staff Emeritus
Ok well its $\frac{at^2}{2}$ just for clarification. If you tell me the result it might help because its a tricky quadratic equation to solve.

6. Jan 6, 2008

1. a) 7.5 s
b) 3.8 m/s

2. a) 15s

7. Jan 6, 2008

### foxjwill

You can possibly use that to simplify the quadratics, but for some reason I had thought you were solving for distance. Oops.

8. Jan 6, 2008

### Kurdt

Staff Emeritus
Well for part 1 a) I got a similar result. I used 7.45 and it was fairly accurate. If you use 7.45 with $s = ut$ you will get, 23.095m and if you use it with $1/2 at^2$ you obtain 13.825 which when summed gives 36.92.

All I can imagine is that you haven't reported the answer to the sufficient accuracy required.

9. Jan 6, 2008

### vertciel

Thank you for your replies, foxjwill and kurdt. I got the answers!