# Homework Help: How long until she passes you?

Tags:
1. Sep 25, 2016

### Mohamed Drahaman

1. The problem statement, all variables and given/known data
You leave a spotlight with an acceleration of 4.9m/s2[fwd] until reaching a speed of 49m/s. A police officer took off from the same intersection 2sec after. She accelerates at 9.8m/s2. How long after this does she pass you? How fast is she going at this time?

2. Relevant equations

3. The attempt at a solution
4.9t2 = 9.8 x (t-2)2

2. Sep 25, 2016

### Staff: Mentor

What type of equations are relevant to the problem? Can you list them?

3. Sep 28, 2016

### victorhugo

Silly question: Does this require calculus and did he write his equations in calculus?

I haven't done calculus yet..

4. Sep 28, 2016

### Staff: Mentor

No calculus required. Just basic kinematic equations. The OP might have made things more clear if he'd availed himself of the x2 icon in the edit panel to create superscripts. It's can be hard to tell sometimes if a '2' is meant to be a constant multiplying something or an exponent.

5. Sep 28, 2016

### victorhugo

Okay, let me attempt it:

at a = 4.9 m/s /s there is a constant acceleration in the + direction until v = 49 m/s
a = m/s /s & v=m/s
a=Δv/t
Thus cancel to get time =
10 s

r = ut + 1/2a t^2
r = 1/2 a t^2 = 245m

( i completely forgot about taking off 2s after)

at t = 2 we can find velocity and distance, but from here i do have a clue how we can find t for when r1 = r2
(if this is how i should do it)

Last edited: Sep 28, 2016
6. Sep 28, 2016

### Staff: Mentor

In your work you didn't account for the 2 second delay before the cop started to accelerate. Also, if it does take the cop 50 seconds to reach the car, the car will have long ago entered its constant cruise speed phase of its motion. That would complicate things...

1. Not really. Unless you're not interested in great accuracy, then you might do it graphically. The area under a velocity versus time graph is distance.
2. Given that v = at, just integrate with respect to t. The integration constant is the initial velocity.

7. Sep 28, 2016

### Staff: Mentor

This would be correct if the police officer caught up with you in under 10 sec. Check this possibility by solving the equation for t to see if that is actually the case.

8. Sep 28, 2016

### victorhugo

I completely forgot, mb. Also, I edited the question instead of replying. Exausted brain doesn't work very well...
Anyway, my edit was:

Okay, let me attempt it:

at a = 4.9 m/s /s there is a constant acceleration in the + direction until v = 49 m/s
a = m/s /s & v=m/s
a=Δv/t
Thus cancel to get time =
10 s

r = ut + 1/2a t^2
r = 1/2 a t^2 = 245m

( i completely forgot about taking off 2s after)

at t = 2 we can find velocity and distance, but from here i do have a clue how we can find t for when r1 = r2
(if this is how i should do it)

9. Sep 28, 2016

### Staff: Mentor

Shouldn't we give the OP a chance to respond?

10. Sep 28, 2016

### victorhugo

I want to play too (:

11. Sep 28, 2016

### Staff: Mentor

You're probably right. But then again, the OP made the one post to start this thread three days ago and hasn't resurfaced.