A motorist travels at a constant speed of 37.0 m/s through a school zone, exceeding the posted speed limit. A policeman waits 7.0 s before giving chase at an acceleration of 3.2 m/s2.
I have solved that it will take 36 seconds for the cop to catch up to the car. However, I am not sure how to set up the equation for solving for distance if both cars have different accelerations. Any detailed help would be very much appreciated. Thank you in advance.
if the motorist is going 37,
distnace=37*time
where the time for the car is 36+7, right?
So if I did my calculation right distance should equal 37 times 43 (36+7) which would equal 1591. Correct?
t=23.78+7=30.78s
x=1046.75m
To solve this problem, we need to first understand the equations of motion.
For an object moving with a constant acceleration, such as the policeman's car, we can use the following equation:
d = ut + (1/2)at^2
Where:
- d is the distance traveled
- u is the initial velocity
- a is the acceleration
- t is the time
For the motorist's car, it is traveling at a constant speed, which means it has zero acceleration. Therefore, its equation of motion simplifies to:
d = ut
Now, let's break down the problem step by step:
1. First, find the initial velocity (u) for both the motorist's car and the policeman's car. We are given that the motorist's car is traveling at a constant speed of 37.0 m/s. As the policeman waits for 7.0 seconds before giving chase, his initial velocity is 0 m/s.
2. Now, let's calculate the time it takes for the policeman to catch up with the motorist. We can use the equation:
t = (v - u) / a
Where:
- v is the final velocity (which will be the same for both vehicles since they meet at the point of catchup)
- u is the initial velocity of the policeman's car
- a is the acceleration of the policeman's car
Substituting the given values, we get:
t = (37.0 m/s - 0 m/s) / 3.2 m/s^2
t = 11.5625 seconds
3. Finally, we can find the distance traveled by both the motorist's car and the policeman's car using their respective equations of motion.
For the motorist's car:
d_motorist = (37.0 m/s) * 11.5625 s
d_motorist = 427.34375 m
For the policeman's car:
d_policeman = (0 m/s) * 11.5625 s + (1/2) * (3.2 m/s^2) * (11.5625 s)^2
d_policeman = 209.0625 m
Therefore, it will take approximately 36 seconds for the policeman to catch up to the motorist, and they will meet at a distance of approximately 427.34 meters from the starting point.