A speeder passes a parked police car at a
constant speed of 23.9 m/s. At that instant,
the police car starts from rest with a uniform
acceleration of 2.21 m/s
2
.
How much time passes before the speeder
is overtaken by the police car?
Answer in units of s
How far does the speeder travel before being
overtaken by the police car?
Answer in units of m
X1 = 23.9 t
X2 = 1,105 t^2
Set X1 = X2 and solve for t
Then use that t and either equation for the location where they meet.
To find the time it takes for the police car to overtake the speeder, we need to find the time it takes for the police car to catch up to the speeder. We can use the equation for motion with uniform acceleration:
v = u + at,
Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time
For the speeder:
u = 23.9 m/s (since the speeder is already moving at a constant speed)
a = 0 m/s^2 (since the speeder maintains a constant speed)
For the police car:
u = 0 m/s (since the police car starts from rest)
a = 2.21 m/s^2 (as given in the problem)
Now we can solve for the time it takes for the police car to overtake the speeder.
For the speeder:
v = 23.9 m/s
t1 = (v - u) / a
t1 = (23.9 m/s - 23.9 m/s) / 0 m/s^2
t1 = 0 s
The time obtained is 0 s, which means the speeder is not moving relative to the police car when the police car starts from rest.
For the police car:
v = 23.9 m/s (since the speeder and police car will have the same velocity when the police car overtakes the speeder)
t2 = (v - u) / a
t2 = (23.9 m/s - 0 m/s) / 2.21 m/s^2
t2 = 10.859 s
Therefore, it takes approximately 10.859 seconds for the police car to overtake the speeder.
To find how far the speeder travels before being overtaken by the police car, we can use the equation:
s = ut + 0.5at^2,
Where:
- s is the displacement
- u is the initial velocity
- a is the acceleration
- t is the time
For the speeder:
u = 23.9 m/s
a = 0 m/s^2
t = 10.859 s (as calculated earlier)
s1 = ut + 0.5at^2
s1 = (23.9 m/s)(10.859 s) + 0.5(0 m/s^2)(10.859 s)^2
s1 ≈ 260.651 m
Therefore, the speeder travels approximately 260.651 meters before being overtaken by the police car.