Speeding Car
A car with velocity V was caught speeding by a police car who started from rest. The police car accelerates as velocity a. How long will it take for the police car to catch up to the speeding car?
I think I simply set up both of the equations for the total distance they will travel from the point the car was caught by cop. However, how to I begin this problem to get it all set up?
Answer
You have the exact right idea on how to solve the problem. You are going to want to start with basic kinematic equations for each car. Since the speeding car is moving at a constant speed, you get the equation
x_c = v*t
The police care starts at rest and accelerates, so you will want to use
x_p = a/2*t^2
Since no distances were specified, I am assuming that both cars start at the same point. If that is not the case you will have to add in the positions at t=0.
The point that the police catches the car is when their cars are at the same point (unless specified otherwise just ignore that cars aren't points) and you solve for t. You will have to be careful, the cars will be at the same point at two different times, so be careful in deciding which one.
Customer support service by UserEcho
You have the exact right idea on how to solve the problem. You are going to want to start with basic kinematic equations for each car. Since the speeding car is moving at a constant speed, you get the equation
x_c = v*t
The police care starts at rest and accelerates, so you will want to use
x_p = a/2*t^2
Since no distances were specified, I am assuming that both cars start at the same point. If that is not the case you will have to add in the positions at t=0.
The point that the police catches the car is when their cars are at the same point (unless specified otherwise just ignore that cars aren't points) and you solve for t. You will have to be careful, the cars will be at the same point at two different times, so be careful in deciding which one.