Question

A speeder passes a parked police car at a constant velocity of 30.0 m/s. The police car starts from rest with a uniform acceleration of 2.44 m/s^2. a) How much time passes before the police car overtakes the speeder? b) How far does the speeder get before being overtaken by the police car?

Answer 1

Take x and y as distance and time and form 2 equations for the police and the speeder.

Answer 2

For a car moving at a constant velocity, the distance traveled at any point in time will be: \[\large d= vt=(30m/s)t\]
For a car starting at 0m/s with a constant acceleration of 2.44m/s^2, the distance at any point in time will be:\[\large d= \cancel{v_0t}+\frac{1}{2}at^2=\cancel{0m/s}+\frac{1}{2}(2.44m/s^2)t^2\]To get the time required for the police car to catch up to the car, simply set these equations equal to each other and solve for t:\[\large (30m/s)t=\frac{1}{2}(2.44m/s^2)t^2\]Since it's quadratic, you'll end up with two values for t: \[\large t=0s \space and \space t=24.59s\]Thinking about this you should see that t=0 represents the point at which the car first passes the police car. Therefore t=24.59s must be the time it took for the police car to catch back up to the car.

Now that you know the time, you can plug that into one of the original equations and determine how far the speeder gets before the police car catches up:\[\large d=(30m/s)(24.59s)=737.7m\]

Answer 3

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