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?

See your 9:32am post.

a.) 25.6m/s

60km/h*1km/1000m=6000m/h

6000m/h*1h/3600=16.6m/s

v=Vo+at
v=16.6+(2.22)(4.00)=25.48 m/s

a)60km/h*1km/1000m=6000m/h

6000m/h*1h/3600=16.6m/s

v=Vo+at
v=16.6+(2.22)(4.00)=25.48 m/s

b)

To solve this problem, we can use the equations of motion and some basic kinematic principles.

a) To find the policeman's speed in m/s, we can use the equation:
v = u + at

where:
v = final velocity
u = initial velocity (which is 60 km/h in this case)
a = acceleration (2.22 m/s^2)
t = time taken (4.00 s)

Converting the initial velocity from km/h to m/s:
u = (60 km/h) x (1000 m/1 km) x (1 h/3600 s) ≈ 16.67 m/s

Now we can substitute the values into the equation and solve for v:
v = 16.67 m/s + (2.22 m/s^2) x (4.00 s)
v = 16.67 m/s + 8.88 m/s
v ≈ 25.55 m/s

Therefore, the policeman's speed is approximately 25.55 m/s.

b) To find the police car's speed after 4.00 s, we can use the same equation:
v = u + at

Using the same initial velocity and acceleration:
u = 16.67 m/s
a = 2.22 m/s^2
t = 4.00 s

Substituting the values:
v = 16.67 m/s + (2.22 m/s^2) x (4.00 s)
v = 16.67 m/s + 8.88 m/s
v ≈ 25.55 m/s

So, after 4.00 seconds, the police car's speed is approximately 25.55 m/s.

To convert this speed from m/s to km/h:
v (km/h) = v (m/s) x (3600 s/1 h) x (1 km/1000 m)
v (km/h) ≈ 25.55 m/s x 3600/1000 ≈ 91.98 km/h

Therefore, the police car's speed after 4.00 s is approximately 25.55 m/s or 91.98 km/h.

c) If the speeder has a constant velocity, its speed will remain the same over time, so it will have a straight line on the speed vs. time graph. The slope of this line will be zero.

For the police car, its speed increases at a steady rate of 2.22 m/s^2, so its speed vs. time graph will be a straight line with a positive slope.

This graph can be visualized as follows:
- The x-axis represents time (in seconds).
- The y-axis represents speed (in m/s).
- The speeder's line will be a horizontal line along the y-axis at the speed of the speeder.
- The police car's line will start at the initial speed of 16.67 m/s and continue with a straight line with a positive slope, representing its acceleration of 2.22 m/s^2.

Please let me know if anything is unclear or if you have any further questions.