How Does a Police Car Catch Up to a Speeding Vehicle?

[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?

Homework Equations

For the Police:

V_p = ∫a_0p(dt) → a_0pt + v_0p

X_p = ∫V_p(dt) → 0.5a_0pt² + v_0pt + X_0p

The Attempt at a Solution

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

X_M is d ∴ d = 0.5a_0pt² + v_0pt + X_0p

However, the v_0pt is = 0 because the cop car is initially at rest. X_0p is also at rest because the placement of the cop car is at the origin.

∴ d = 0.5a_0pt²

Solving for t gives: t = (2d/a_0p)^0.5. You can discard the negative value for t.

[b.] For part B, the distance the cop car has traveled can be calculated simply by plugging in the value of t into 0.5a_0pt² + v_0pt + X_0p

Once again, the v_0pt is = 0 because the cop car is initially at rest. X_0p is also at rest because the placement of the cop car is at the origin.

∴X_P = 0.5a_0p(2d/a_0p)^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, V_P = a_0pt

= a_0p(2d/a_0p)^0.5)²

 

[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?

 

[Metamorphose]

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

 

[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.

 

[Metamorphose]

Oh! I forgot to integrate my V₀, which should have given me X_M = V_0Mt + X_0M.

Which should have been:

V_0Mt + d = 0.5a_0pt² + v_0pt + X_0p

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

 

[Metamorphose]

And when this quadratic is solved:

t = (2V_0M + (4V_0m² + 8da_0p)^0.5)/2a_0P.

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

 

[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.

 

[Metamorphose]

is the answer I provided above correct?

 

[LawrenceC]

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