Solving Kinematics: Police Car Chasing Speeders

[Rikochet]

Homework Statement

A police car stopped at a set of lights has a speeder pass it at 100 km/h. If the police car can accelerate at 3.6 m/s squared

a) how long does it take to catch the speeder

b) how far would the police car have to go before it catches the speeder?

c) what would its speed be when it caught up with the car? Is this speed reasonable?

Homework Equations

not sure about these but:

v2 = v1 + a(delta)t

d = speed over time.

The Attempt at a Solution

At first I thought I just needed to calculate how long it would take for the police car to get to 100 km/h, but then I realized that the speeder is going 100km/h. I'm just really confused because there are two objects that you have to keep track of instead of just one. I have no clue where to start on this question.

 

[ryan88]

Since you know that you are trying to find the time taken for the police car to catch the other car, you know that when he does, their displacements are going to be the same. So if you get an equation for the displacement of each car, you can equate them to each other to find the time. So do you know a kinematic equation that relates displacement, initial velocity, acceleration and time?

 

[tomcue]

I would approach this problem by first defining the variables involved. In this case, we have the speed of the speeder (100 km/h), the acceleration of the police car (3.6 m/s squared), and the initial speed of the police car (0 m/s). We also need to consider the distance between the two vehicles, which we can assume is initially some value "d" that we do not know.

Next, I would use the kinematic equations to solve for the unknown variables. For part (a), we can use the equation v2 = v1 + a(delta)t, where v2 is the final velocity of the police car (which we want to be 100 km/h), v1 is the initial velocity (0 m/s), a is the acceleration (3.6 m/s squared), and (delta)t is the time it takes for the police car to catch up to the speeder. Rearranging the equation, we get (delta)t = (v2 - v1)/a. Plugging in the values, we get (delta)t = (100 km/h - 0 m/s)/(3.6 m/s squared) = 27.8 seconds.

For part (b), we can use the equation d = v1(delta)t + 1/2a(delta)t^2, where d is the distance traveled by the police car (which we do not know), v1 is the initial velocity (0 m/s), a is the acceleration (3.6 m/s squared), and (delta)t is the time it takes for the police car to catch up to the speeder (27.8 seconds). Rearranging the equation, we get d = 1/2a(delta)t^2 = 1/2(3.6 m/s squared)(27.8 seconds)^2 = 1388.8 meters.

For part (c), we can use the equation v2^2 = v1^2 + 2ad, where v2 is the final velocity of the police car (which we want to be 100 km/h), v1 is the initial velocity (0 m/s), a is the acceleration (3.6 m/s squared), and d is the distance traveled by the police car (1388.8 meters). Rearranging the equation, we get v2 = sqrt(v1^2 + 2ad) = sqrt(0 m