A speeder passes a parked police car at 30 m/s and continues on, obliviously maintaining a constant velocity. At that instant, the police car starts from rest with a uniform acceleration of 2.44 m/s2.
a) How much time passes before the officer catches up to the speeder?
b) How far does the police car travel in that time?
c) How fast is the police car going when it catches up with the speeder?
What would the equations be for each of these questions
s=2u^2/a
s=2*30^2/2.44
s=737.7
s=ut
t=s/u
t=737.7/30
t=24.59sec
To solve this problem, we can use the equations of motion. The relevant equations for this scenario are:
1) v = u + at - This equation relates initial velocity (u), final velocity (v), acceleration (a), and time (t).
2) s = ut + (1/2)at^2 - This equation relates displacement (s), initial velocity (u), acceleration (a), and time (t).
Now, let's use these equations to solve each part of the problem:
a) How much time passes before the officer catches up to the speeder?
In this situation, the speeder's velocity remains constant at 30 m/s while the police car starts from rest with an acceleration of 2.44 m/s^2. To find the time it takes for the police car to catch up to the speeder, we need to equate the velocities.
Let v_p be the velocity of the police car when it catches up to the speeder, and v_s be the speeder's velocity. We can write the equation as:
v_p = v_s
Now we substitute the values:
v_p = 30 m/s (speeder's velocity)
For the police car, we have:
v_p = u + at
Since the police car starts from rest (u = 0), we have:
v_p = at
Substituting the acceleration value, we get:
2.44t = 30
Solving for t, we divide both sides by 2.44:
t = 30 / 2.44
t ≈ 12.3 seconds
So, it takes approximately 12.3 seconds for the officer to catch up to the speeder.
b) How far does the police car travel in that time?
To find the distance traveled by the police car during this time, we can use the equation:
s = ut + (1/2)at^2
Since the initial velocity (u) of the police car is 0, the equation reduces to:
s = (1/2)at^2
Substituting the given values:
s = (1/2)(2.44)(12.3)^2
s ≈ 184.3 meters
Therefore, the police car travels approximately 184.3 meters in that time.
c) How fast is the police car going when it catches up with the speeder?
To find the final velocity (v_p) of the police car when it catches up to the speeder, we can use the equation:
v_p = u + at
Again, since the initial velocity (u) is 0, the equation simplifies to:
v_p = at
Substituting the values:
v_p = 2.44(12.3)
v_p ≈ 30 meters per second (same as the speeder's velocity)
Therefore, the police car is going approximately 30 m/s when it catches up with the speeder.