How do I go about drawing a graph for this problem?

Question

A police car is at rest when a speeder traveling at a constant 110km/h passes by. The police car takes off in hot pursuit the instant the speeder passes by. It catches up to the speeder in 700m, maintaining a constant acceleration.

The graph is supposed to be position versus time.

Answer 1

An interesting graph to consider is:
separation of the cars versus time

at time = 0, the cars are zero distance apart. The problem is, the police car is at rest, and the speeder is moving at 110 km/hr

so, let's look at some equations of motion for the two cars:

The velocity of the speeder is
vs(t)=110 km/he
the velocity of the police car is
VP(t)=a*t

the location of the speeder is
ls=110 *t (t is in hours)

the location of the police car is
lp=.5*a*t^2 where t is in hours and a is m/hr^2

let's look at ls-lp, which is the separation between the two cars

ls-lp=110 *t -.5*a*t^2

note that when ls-LP = 0, there are two points, t=0, and the point where the police car overtakes the speeder

so t^2=2*110/a

but what is a?
a=220/t^2

we know that when lp=700 m, the overtake occurs, and since we know that the speeder reaches the overtake in time t,
then 110 *t=700
t=700/110
you can now solve for a and plot the distance of separation.

You can also plot the speed of the speeder versus time, and the speed of the police car versus time.

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