Question

A 250 m long street race is staged between two cars. Car B is given a 10 m head start (that is, it begins 10 m in front of Car A). At the same time, both cars begin accelerating from rest. Car A has an acceleration of 4.25 m/s2, while Car B has an acceleration of 3.95 m/s2. The cars keep racing past the 250 m distance marking the end of the race.

a. Which car wins the race? What is their winning time?
b. Eventually, Car A will catch up to Car B - where and when?
c. At the point where Car A catches up to Car B, what are their velocities?

Answer 1

To solve for when Car A will catch up to Car B in the street race, we apply kinematic equations for motion due to acceleration. The winner, winning time, catch-up point, and velocities are all determined through equations involving acceleration, time, and initial velocity.

Step-by-step explanation:

In the scenario where a 250 meter long street race is staged between two cars, with Car B given a 10 meter head start, we can determine the winner, the catch-up point, and the velocities at that point using kinematic equations of motion. Both cars accelerate from rest, Car A at 4.25 m/s² and Car B at 3.95 m/s².

To find the winner and their winning time, we use s = ut + (1/2)at², where 's' is displacement, 'u' is initial velocity, 't' is time, and 'a' is acceleration. With Car A's starting displacement being 0 m and Car B's 10 m, we set up two equations and solve for 't' when 's' is 250 m.

To determine when and where Car A will catch up to Car B, we equate the two distances traveled by the cars relative to their respective accelerations. This involves solving for 't' when the distances are equal, accounting for the head start.

Once we know the time at which Car A catches Car B, we can calculate their velocities using v = u + at. Since both cars started from rest, their velocity is simply at.