A speeder passes a parked police car at a constant speed of 27.3 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.55 m/s2 .
a) How much time passes before the speeder is overtaken by the police car?
b) How far does the speeder travel before being overtaken by the police car?
s=2u^2/a
s=2*27.3^2/2.55
s=584.54
s=ut
t=s/u
t=584.54/27.3= 21.41sec
see related questions below.
To solve this problem, we will use kinematic equations that relate the initial velocity, final velocity, acceleration, time, and displacement of an object.
Let's analyze the situation given in the problem:
For the speeder:
Initial velocity (u) = 27.3 m/s
Final velocity (v) = ?
Acceleration (a) = 0 m/s^2 (since the speeder is moving at a constant speed)
Time (t) = ?
Displacement (s) = ?
For the police car:
Initial velocity (u) = 0 m/s (since it starts from rest)
Final velocity (v) = ?
Acceleration (a) = 2.55 m/s^2
Time (t) = ?
Displacement (s) = ?
a) To find the time it takes for the police car to overtake the speeder, we need to find the time when both the speeder and the police car have the same final velocity. This can be calculated using the equation:
v = u + at
For the speeder:
v = 27.3 m/s (since it is moving at a constant speed)
u = 27.3 m/s
a = 0 m/s^2
t = ?
Replacing the values in the equation, we get:
27.3 = 27.3 + 0t
0t = 0
Therefore, we can see that the time it takes for the speeder to be overtaken by the police car is undefined, as the speeder will always have a greater velocity. The police car will never catch up with the speeder.
b) As we found in part a), the police car cannot overtake the speeder. Therefore, the distance traveled by the speeder before being overtaken is also undefined.
In summary:
a) The time it takes for the speeder to be overtaken by the police car is undefined.
b) The distance traveled by the speeder before being overtaken by the police car is also undefined.