A motorist travels at a constant speed of 34.0 m/s through a school zone; exceeding the posted speed limit. A policeman, waits 7.0 s before giving chase at an acceleration of 3.9 m/s2.
(a) Find the time required to catch the car, from the instant the car passes the policeman.
(b) Find the distance required for the policeman to overtake the motorist.
3Your answer is incorrect.
IT is difficult to know where the policeman started, so I assume at some point the motorist passed.
They both traveled the same distance.
d= 34*t
d=1/2 3.9 (t-7)^2
set them equal, and solve for time.
then solve for distance.
To solve this problem, we can use the equations of motion. Let's break down the problem and find the solutions step by step.
(a) Find the time required to catch the car, from the instant the car passes the policeman.
First, let's find the initial velocity of the policeman when he starts chasing. The policeman waits for 7.0 s, during which the car continues to travel at a constant speed. Therefore, the initial velocity of the policeman will be 0 m/s.
We'll use the equation of motion:
v = u + at
where:
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
Since the final velocity of the car is 34.0 m/s and the initial velocity of the policeman is 0 m/s, we can rearrange the equation to solve for time (t):
t = (v - u) / a
t = (34.0 m/s - 0 m/s) / (3.9 m/s^2)
t ≈ 8.72 s
Therefore, it will take approximately 8.72 seconds for the policeman to catch up to the car.
(b) Find the distance required for the policeman to overtake the motorist.
To find the distance required for the policeman to overtake the motorist, we'll use the equation of motion:
s = ut + (1/2)at^2
where:
s is the distance
u is the initial velocity
t is the time
a is the acceleration
Since the initial velocity of the policeman is 0 m/s, we can simplify the equation to:
s = (1/2)at^2
Plugging in the values, we get:
s = (1/2)(3.9 m/s^2)(8.72 s)^2
s ≈ 152.27 m
Therefore, the distance required for the policeman to overtake the motorist is approximately 152.27 meters.
Please note that these calculations assume constant acceleration and do not take into account any other factors that may affect the outcome in a real-world scenario.