Motorcycle trooper overtaking a car.

[DatAshe]

Homework Statement

A car traveling at a constant speed of 51.5 m/s passes a trooper on a motorcycle hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets out from the billboard to catch the car, accelerating at a constant rate of 2.00 m/s2. How long does it take her to overtake the car? Solve this problem by a graphical method. On the same graph plot position versus time for the car and the police officer. (Do this on paper. Your instructor may ask you to turn in this work.)

From the intersection of the two curves read the time at which the trooper overtakes the car.
s

Homework Equations

a=(v-v_o)/t

d/Δt=constant speed

The Attempt at a Solution

I don't have anything good to put here. I have no clue what I am doing on this question. I also have a question about a race car overtaking another race car and I have no idea of what to do on that one either. I would really appreciate a conceptual explanation over an answer, because I want to know how to do it for the future. Thank you!

 

[szynkasz]

You need "distance vs time" equations for car and trooper.

 

[DatAshe]

What would an example of a distance vs time equation be? Like, speed=d/deltat or acceleration=deltax/deltat or something else?

 

[szynkasz]

You have motion with constant speed and motion with constant acceleration. There are formulas for distance vs time for both.

 

[azizlwl]

It is velocity vs. time graph.
As you see the car is traveling at constant speed.
For the trooper for first second, his velocity is zero.
Then he accelerates at constant rate.
Acceleration means the Δv/Δt, the ratio of the change of velocity over change of time.
Since it is constant, it must be a straight line.

The time taken for the trooper to overtake the car is the point where he is same displacement/location as the car.

Displacement of the car is just a rectangle area and the trooper is a triangular area(velocity x time).

Then on the same graph you can plot position versus time for the car and the police officer