Final Exam Review: Police Car

1. May 6, 2012

Metamorphose

1. While driving on the highway at a constant speed v0, significantly above the speed limit,
you pass in front of a parked police car without noticing it. After a few seconds, during which time you have moved a distance d, the police car starts chasing you with a constant acceleration a0. Give your answers in terms of v0, d, and a0 and check the units of your answers. Assuming that you keep moving at a constant speed, and that the highway can be considered straight:

[a] How long will it take for the police car to reach you?

[b.] How far has the police car traveled when it reaches you?

[c] What is the speed of the police car when it reaches you?

2. Relevant equations

For the Police:

Vp = ∫a0p(dt) → a0pt + v0p

Xp = ∫Vp(dt) → 0.5a0pt2 + v0pt + X0p

3. The attempt at a solution

[a] The police car will reach me when our positions are the same, ∴ XM = XP

XM is d ∴ d = 0.5a0pt2 + v0pt + X0p

However, the v0pt is = 0 because the cop car is initially at rest. X0p is also at rest because the placement of the cop car is at the origin.

∴ d = 0.5a0pt2

Solving for t gives: t = (2d/a0p)0.5. You can discard the negative value for t.

[b.] For part B, the distance the cop car has travelled can be calculated simply by plugging in the value of t into 0.5a0pt2 + v0pt + X0p

Once again, the v0pt is = 0 because the cop car is initially at rest. X0p is also at rest because the placement of the cop car is at the origin.

∴XP = 0.5a0p(2d/a0p)0.5)2

which gives d as an output. I am not quite sure if this is the correct answer or not as I have no solutions manual to compare my answer to.

[c.] For part C, the speed of the police car when it reaches me should be equivalent to my speed, which is V0

therefore, VP = a0pt

= a0p(2d/a0p)0.5)2

Last edited: May 6, 2012
2. May 6, 2012

LawrenceC

For part A: You have not considered the total distance covered by the speeding car. Cop does not start until speeder has a distance, d, on him. But what about the distance speeder covers while the cop accelerates?

3. May 6, 2012

Metamorphose

I'm not quite sure how to incorporate that into this. I'm sorry

4. May 6, 2012

LawrenceC

OK, no problem.

"Assuming that you keep moving at a constant speed,...."

The distance you will have traveled during that time is Vyou * t, where the t is the same time the police car accelerates until he catches you. That adds additional distance to the equation.

5. May 6, 2012

Metamorphose

Oh! I forgot to integrate my V0, which should have given me XM = V0Mt + X0M.

Which should have been:

V0Mt + d = 0.5a0pt2 + v0pt + X0p

At which point, solving for t should be solved using a quadratic equation?

6. May 6, 2012

Metamorphose

And when this quadratic is solved:

t = (2V0M + (4V0m2 + 8da0p)0.5)/2a0P.

The negative solution for time t can be discarded. This also alters my answers to b and c

7. May 6, 2012

LawrenceC

You could solve by 'completing the square' as it was called when I was in school. Or you could use the quadratic formula which is derived by completing the square.

8. May 6, 2012

Metamorphose

is the answer I provided above correct?

9. May 6, 2012

LawrenceC

Yes but I would simplify it a bit by putting the 2 inside the radical and removing it from the first term.