How long until she passes you?

[Mohamed Drahaman]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

4.9t2 = 9.8 x (t-2)2

 

[gneill]

You'll need to show more of your attempt and describe what you are trying to do there. What does your equation represent?

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

 

[victorhugo]

You'll need to show more of your attempt and describe what you are trying to do there. What does your equation represent?

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

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

I haven't done calculus yet..

 

[gneill]

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.

 

[victorhugo]

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.

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)

 

[gneill]

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. is there another way to solve using simpler maths
2. can you give me a logical explanation to how that formula works? e.g I understand velocity and acceleration because of the definitions, so I can intuitively solve for each.

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.

 

[Chestermiller]

Homework Statement

The Attempt at a Solution

4.9t2 = 9.8 x (t-2)2

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.

 

[victorhugo]

In your work you didn't account for the 2 second delay before the cop started to accelerate.

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)

 

[Chestermiller]

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

 

[victorhugo]

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

I want to play too (:

 

[gneill]

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

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