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.

answer; (arch); good luck

radiation

explanation:

heat is transmitted by radiation