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?
a. Vo=60km/h = 60,000m/3600s=16.67m/s.
b. V=Vo + at = 16.67 + 2.22*4 =25.55m/s.
V = 0.02555km/s * 3600s/h = 92 km/h.
To solve this problem, we can use the equations of motion:
a) To find the speed of the policeman in m/s, we need to convert 60 km/h to m/s. We know that 1 km/h is equal to 0.2778 m/s. Therefore,
Speed of the policeman = 60 km/h * 0.2778 m/s = 16.668 m/s
b) To find the speed of the police car after 4.00 s, we can use the equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Initial velocity (u) of the police car is 16.668 m/s (as calculated in part a), acceleration (a) is 2.22 m/s^2, and time (t) is 4.00 s. Plugging in these values into the equation:
v = 16.668 m/s + (2.22 m/s^2 * 4.00 s)
v = 16.668 m/s + 8.88 m/s
v = 25.548 m/s
To convert the final velocity from m/s to km/h:
Final velocity (km/h) = v * (3600 / 1000) = 25.548 m/s * 3.6 = 91.9728 km/h
Therefore, the speed of the police car after 4.00 s is approximately 25.548 m/s or 91.9728 km/h.
c) Since the speeder has a constant velocity, the speed vs. time graph for the speeder will be a horizontal straight line because the speed remains constant over time.
On the other hand, the police car is accelerating at a steady rate, so the speed vs. time graph for the police car will be an upward sloping straight line indicating an increase in speed over time.
To answer these questions, we need to use the kinematic equations of motion.
a) To find how fast the policeman was traveling in m/s, we need to calculate his final velocity using the equation:
vf = vi + at
Where:
- vf is the final velocity,
- vi is the initial velocity,
- a is the acceleration, and
- t is the time.
In this case, the initial velocity (vi) is 60 km/h, which needs to be converted to m/s. We know that 1 km/h is equivalent to 0.2778 m/s. Therefore, 60 km/h would be 16.67 m/s.
Plugging these values into the equation:
vf = 16.67 m/s + (2.22 m/s^2)(4.00 s)
vf = 16.67 m/s + 8.88 m/s
vf = 25.55 m/s
Therefore, the policeman was traveling at 25.55 m/s.
b) To find the final velocity of the police car after 4.00 s, we can use the same equation:
vf = vi + at
In this case, the initial velocity (vi) is 0 m/s (since the car was initially stationary), and the acceleration (a) is still 2.22 m/s^2. Plugging in the values:
vf = 0 m/s + (2.22 m/s^2)(4.00 s)
vf = 0 m/s + 8.88 m/s
vf = 8.88 m/s
To convert this velocity from m/s to km/h, we multiply by 3.6 (because 1 m/s is equivalent to 3.6 km/h):
vf = 8.88 m/s * 3.6 km/h
vf = 31.96 km/h
Therefore, the police car is traveling at 8.88 m/s or 31.96 km/h after 4.00 s.
c) Since the speeder has a constant velocity, their speed vs. time graph would be a straight horizontal line, indicating constant speed. The police car's speed vs. time graph would start at 0 m/s and increase linearly with time at a slope corresponding to the rate of acceleration (2.22 m/s^2). After 4.00 s, the police car's speed would intersect with the speed of the speeder and maintain the same constant speed as the speeder.