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) To find the speed of the policeman in m/s, we need to first convert the speed from km/h to m/s.
Given:
Policeman's initial speed = 60 km/h
Conversion:
1 km = 1000 m
1 hour = 3600 seconds
Speed in m/s = (60 km/h) * (1000 m/km) / (3600 s/h) = 16.67 m/s
Therefore, the policeman was traveling at a speed of 16.67 m/s.
b) To find the speed of the police car after 4.00 s, we need to use the equation of motion:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
Initial velocity, u = 16.67 m/s
Acceleration, a = 2.22 m/s^2
Time, t = 4.00 s
Using the equation of motion:
v = u + at
v = 16.67 m/s + (2.22 m/s^2) * 4.00 s
v = 16.67 m/s + 8.88 m/s
v = 25.55 m/s
To convert the final speed to km/h:
Speed in km/h = (25.55 m/s) * (3600 s/h) / (1000 m/km) = 92.0 km/h (approximately)
Therefore, after 4.00 s, the police car is traveling at a speed of 25.55 m/s or 92.0 km/h.
c) If the speeder has a constant velocity, the speed vs. time graph would show a horizontal line at the speed of the speeder. Let's assume the speeder is traveling at a constant speed of 16.00 m/s.
On the graph, the speed of the speeder would be a straight, horizontal line at a height of 16.00 m/s for all values of time.
The police car, on the other hand, starts at an initial speed of 16.67 m/s and accelerates at a constant rate of 2.22 m/s^2. Therefore, the speed vs. time graph for the police car would show a straight line with a positive slope. The slope represents the rate of acceleration.
Note: Without specific values for the speeder's speed, it is not possible to provide accurate points on the graph or determine the time when the police car catches up with the speeder.
To solve this problem, we'll break it down into parts.
a) To find the speed of the policeman in m/s, we need to determine the final velocity of the policeman's vehicle after accelerating for 4.00 s. We can do this by using the formula:
v = u + at
Where:
v = final velocity
u = initial velocity (speed of the policeman traveling at 60 km/h)
a = acceleration (2.22 m/s^2)
t = time (4.00 s)
First, we need to convert the initial velocity of the policeman from km/h to m/s. To do this, we multiply the speed in km/h by 1000/3600 to convert it to m/s:
Initial Velocity (u) = 60 km/h * (1000 m/3600 s) = 16.67 m/s
Now we can substitute our values into the formula and solve for v:
v = 16.67 m/s + (2.22 m/s^2 * 4.00 s)
v = 16.67 m/s + 8.88 m/s
v = 25.55 m/s
Therefore, the speed of the policeman is 25.55 m/s.
b) To find the speed of the police car after 4.00 s, we can use the final velocity we calculated in part a). The speed is the magnitude of the velocity, so it is the same value.
Speed after 4.00 s = 25.55 m/s
To convert this value back to km/h, we multiply it by 3600/1000:
Speed after 4.00 s = 25.55 m/s * (3600 km/1000 m) = 91.98 km/h
Therefore, the speed of the police car after 4.00 s is approximately 91.98 km/h or 25.55 m/s.
c) Since the speeder has a constant velocity, their acceleration is zero. On a speed vs. time graph, this means the speeder's line will be horizontal, indicating a constant speed.
The police car, on the other hand, is initially traveling at 60 km/h (16.67 m/s) and then accelerating at a constant rate. So the police car's speed vs. time graph will start with a horizontal line (indicating a constant speed), and then a straight diagonal line, showing an increase in speed due to acceleration.
The specific slope of the graph for the police car will depend on the rate of acceleration.