A speeder passes a parked police car at a
constant speed of 39.2 m/s. At that instant,
the police car starts from rest with a uniform
2
is overtaken by the police car? Answer in units of s
acceleration of 2.77 m/s .
How much time passes before the speeder
To find the time it takes for the police car to overtake the speeder, we can use the equations of motion.
Let's denote the initial velocity of the police car as u1 (which is 0 m/s since it starts from rest) and the final velocity of the police car as v1. The initial velocity of the speeder is v2 (which is 39.2 m/s), and the final velocity of the speeder is u2 (which is also 39.2 m/s since it is moving at a constant speed).
We know that the acceleration of the police car is 2.77 m/s^2, and we need to find the time it takes for the police car to reach the same speed as the speeder.
We can use the equation of motion:
v1 = u1 + a1 * t
Since the initial velocity of the police car is 0 m/s, the equation simplifies to:
v1 = a1 * t
Now, let's solve for the time it takes for the police car to reach the speed of the speeder:
t = v1 / a1
To find v1, we need to find the final velocity of the speeder when the police car overtakes it. Since both the speeder and the police car are moving at the same constant speed of 39.2 m/s, the final velocity v1 is also 39.2 m/s.
Substituting the values into the equation, we get:
t = 39.2 m/s / 2.77 m/s^2
Now we can calculate the value:
t ≈ 14.12 seconds
Therefore, it takes approximately 14.12 seconds for the police car to overtake the speeder.