Question

This police car then moves at a constant speed.

a) will the police car overtake the speeder if the police car moves just as the speeder passes it?
b) How far has each car travelled ?
c) sketch x(t) for each car

Answer 1

The police car will overtake the speeder since it takes less time for the police car to catch up. The distance traveled by the speeder is 1000 m, while the police car travels a distance of 200 m. Sketching x(t) for each car shows a straight line graph for the speeder and a steeper line graph for the police car.

Step-by-step explanation:

In order to answer the questions, we need to determine the distance traveled by each car and the time it takes for the police car to catch up to the speeder.

For the speeding car:

1. Velocity (v) = 40 m/s
2. Time (t) = 25 s
3. Distance (d) = v x t = 40 m/s x 25 s = 1000 m

For the police car:

1. Acceleration (a) = 4 m/s²
2. Time (t) = 20 s
3. Initial velocity (u) = 0 m/s
4. -100 = 0.5a(t^2)
5. Solving for t, we get t = 10 s
6. Using the equation of motion, x = ut + 0.5at^2, we can calculate the distance traveled by the police car: x = 0 + 0.5 x 4 m/s² x (10 s)^2 = 200 m

The police car will overtake the speeder since it takes less time for the police car (10 s) to catch up to the speeder (25 s).

Sketching x(t) for each car, the graph for the speeding car will be a straight line with a slope of 40 m/s and a duration of 25 s. For the police car, the graph will start at 0 and gradually increase with a steeper slope of 4 m/s².