Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance DA beyond the starting line at t=0. The starting line is at x=0. Car A travels at a constant speed vA. Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed vB, which is greater than vA?

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed , which is greater than .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Part A:

              Part B:

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

where

is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is .

The equation will be:

Car B started at the starting line. So, its equation is

Part A: When they meet, both car are at "the same position":

Car B meet with Car A after units of time.

Part B: With the meeting time, we can determine the position they will be:

Since Car B started at the starting line, the distance Car B will be when it passes Car A is units of distance.

molecules are free to separate from each other and will take on the shape of the container. molecules move completely at random and high speeds. they vibrate in a fixed position and do not move around much. molecules expand to the edge of their container.