A policeman travelling 60 km/h spots a speeder ahead , so he accelerates his vehicle at a steady rate of 2.22 m/s^2 for 4.00 s, at which time he catches up with the speeder.
a) How fast was the policeman travelling in m/s?
b) How fast is the police car travelling after 4.00 s? Give answer in both m/s and km/h?
c) If the speeder has a constant velocity, sketch the motion of both the speeder and police car on a speed vs. time graph?
To find the answers to these questions, we can use the equations of motion.
a) To find the initial velocity of the policeman, we need to know the distance he traveled in the 4 seconds. We can calculate this using the equation:
distance = initial velocity × time + 0.5 × acceleration × time^2
The acceleration is given as 2.22 m/s^2, and the time is 4.00 s. We need to solve for the initial velocity. Rearranging the equation, we get:
initial velocity = (distance - 0.5 × acceleration × time^2) / time
Since the policeman caught up with the speeder, the distance traveled by both is the same. Therefore, we can set the distance equal to each other:
60 km/h × (1,000 m/1 km) × (1 h/3,600 s) × 4.00 s = initial velocity × 4.00 s + 0.5 × 2.22 m/s^2 × (4.00 s)^2
Simplifying this equation will give us the initial velocity in m/s.
b) The final velocity of the police car after 4.00 s can be calculated using the equation:
final velocity = initial velocity + acceleration × time
We already have the initial velocity from part (a), and the acceleration is still 2.22 m/s^2. We can plug in the values to find the final velocity.
To convert the final velocity from m/s to km/h, you can use the conversion factor: 1 km/h = 0.2778 m/s.
c) Since the speeder has a constant velocity, its speed vs. time graph will be a horizontal line. The police car's graph will initially be a straight line with positive slope on the speed vs. time graph (indicating it is accelerating), and after 4.00 seconds, it will have a new horizontal line with constant speed matching the speeder's velocity.
To sketch the motion of both the speeder and the police car on a speed vs. time graph, you can plot a horizontal line for the speeder and a diagonal line for the police car. The slope of the police car's line represents its acceleration. After 4.00 seconds, the line will flatten out at the speeder's constant velocity.
By following these steps, you should be able to find the answers to all three parts of the question.