a speeder with unknown velocity passes a police car.the policr car starts to accelerate at a rate of2.2 m/s/s in the direction of the speeder as soon as the speeder passes.the police car reaches the same velocity as the speeder after he travelled 715 m. What is the velocity of the speeder
The velocity attaned by the police car after traveling X = 715 m is
V = sqrt(2aX) = 56.1 m/s
= 202 km/h = 125.5 mph
That must also be the speeder's velocity, since the speeder is not accelerating.
A speeder passes a parked police car at a
constant speed of 150 m/s. At that instant,
the police car starts from rest with a uniform
acceleration of 10 m/s
2
.
How much time passes before the speeder
is overtaken by the police car?
Answer in units of s
To determine the velocity of the speeder, we need to utilize the kinematic equations.
Let's break down the information provided:
1. The police car starts accelerating at a rate of 2.2 m/s^2.
2. The police car reaches the same velocity as the speeder after traveling 715 m.
Now, we can use the following kinematic equation to find the velocity of the speeder:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
s = distance
Since the initial velocity (u) of the speeder is unknown, we can assume it as 0, as it was not mentioned. The final velocity (v) of the speeder will be the same as the final velocity of the police car.
Considering that the police car starts from rest (initial velocity = 0) and reaches the same velocity as the speeder (final velocity = v), we can rewrite the equation as follows:
v^2 = 0^2 + 2(2.2)(715)
Simplifying further:
v^2 = 2(2.2)(715)
v^2 = 3146
v ≈ √3146
v ≈ 56.06 m/s
Therefore, the velocity of the speeder is approximately 56.06 m/s.