Question

ustapha Jones is speeding on the interstate in his Ferrari at 231km/hr when he passes a police car at rest . If the cop accelerates from rest at 15m/s2 starting at the instant Moustapha is next to him, and if Moustapha maintains his speed, how fast is the cop going when he reaches Moustapha

Answer 1

Answer:

461 km/h

Explanation:

In order to solve this problem we must first sketch a drawing of what the situation looks like so we can better visualize it. (See attached picture).

We have two situations there, the first one is Mustapha's car that is traveling at a constant speed of 231km/hr.

The second situation is the police that is accelerating from rest until he reaches Mustapha. (We are going to suppose the acceleration is constant and that he will not stop accelerating until he reaches Mustapha). He has an acceleration of 15m/$s^{2}$.

We want to find what the final velocity of the police is at the time he reaches Mustapha. From this we can imply that the displacement x will be the same for both particles.

So let's model the first situation.

The displacement of Mustapha can be found by using the following equation:

$V_{M}=\frac{x}{t}$

when solving the equation for the displacement x we get that it will be:

$x=V_{M}t$

Now let's model the displacement of the cop. Since the cop has a constant acceleration, we can model his displacement with the following formula.

$x=V_{0}t+\frac{1}{2}at^{2}$

Since the initial velocity of the cop is zero, we can get rid of that part of the equation leaving us with:

$x=\frac{1}{2}at^{2}$

We can now set both equations equal to each other so we get:

$\frac{1}{2}at^{2}=V_{M}t$

When solving this for t, we get that:

$t=\frac{2V_{M}}{a}$  (let's call this equation 1)

Now, we know the cop has constant acceleration, so we can model it with the following formula too:

$a=\frac{V_{f}-V_{0}}{t}$

since the initial velocity of the cop is zero, we can get rid of that here too, so we get the following formula:

$a=\frac{V_{f}}{t}$

when solving for the final velocity, we get that:

$V_{f}=at$  (let's call this equation 2)

when substituting equation 1 into equation 2 we get:

$V_{f}=a(\frac{2V_{M}}{a})$

we can now cancel a leaving us with:

$V_{f}=2V_{M}$

This tells us that the final velocity of the cop will not depend on his acceleration. (This is only if the acceleration is constant all the time) So we get that the final velocity of the cop is:

$V_{f}=2(231km/hr)$

so

$V_{f}=461km/hr$

which is our answer.