A bicycle racer sprints at the end of a race to clinch a...

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 $\mathrm{m} / \mathrm{s}^{2}$ for 7.00 $\mathrm{s}$ . (a) What is his final velocity? (b) The racer continues at this velocity to the finish line. If he was 300 $\mathrm{m}$ from the finish line when he started to accelerate, how much time did he save? (c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

Key Concepts

Time and Distance Relationships

The relationship between time and distance is fundamental in kinematics. It involves understanding how the distance covered by an object relates to its speed or velocity over time. In scenarios with phases of acceleration and constant velocity, calculating the total time taken to cover a given distance or the difference in time between two objects is essential to determine outcomes such as time savings or positional advantages.

Kinematic Equations

Kinematic equations are a set of formulas that relate the variables of motion under uniform acceleration—namely displacement, initial velocity, final velocity, acceleration, and time. These equations are essential for solving problems in mechanics where one or more of these variables are unknown, and they provide a systematic approach to analyzing linear motion scenarios.

Uniform Acceleration

Uniform acceleration refers to a constant rate of change of velocity over time. In problems involving uniformly accelerated motion, the acceleration remains constant, meaning that the equations of motion can be applied directly. This concept is crucial because it allows for the use of standard kinematic equations to calculate final velocity, time, and displacement when acceleration is constant.

Relative Motion

Relative motion involves comparing the motion of two or more objects with respect to each other. In problems where multiple objects (or racers) move with different velocities or accelerations, analyzing their motion relative to one another can reveal differences in displacement, time, or position at a specific moment, which is key to determining advantages or delays.

Transcript

00:01 Part a, we want to find the final velocity.

00:02 We can say that the final velocity is equal to the initial velocity plus the acceleration times time.

00:07 This would be equaling 11 .5 meters per second plus the acceleration of 0 .50 meters per second squared, multiplied by the time of 7 .00 seconds.

00:25 This is equaling 15 .0 meters per second.

00:31 This would be our answer for part a.

00:36 Now for part b, we can see that the expression for time taken when he is running at a constant velocity, we can say t sub c, so t constant would be equal to the distance divided by the velocity.

00:50 This would be equaling 300 meters divided by 11 .5 meters per second.

00:57 This is going to be equaling to 26 .09 seconds.

01:01 We can then say that the here the distance s or the distance we can say delta x traveled by the car would be equalling vx initial times t plus one half times the acceleration and the extraction times t squared and then we're going to solve this would be equalling 11 .5 meters per second this would be multiplied by 7 .00 seconds plus one half of 0 .5 meters per second squared times seven seconds squared.

01:40 And so we have that delta x is equaling 92 .75 meters.

01:47 And so we can say that the time taken by the racer at the constant final velocity to complete the race, we can say t prime would be equaling 300 meters minus 92 .7 5 meters.

02:09 This would be divided by 15 .0 meters per second and this is equaling 13 .82 seconds and so the total time would be equaling 13 .82 plus 7 .00 seconds.

02:38 This is 20, 13 .82 plus 7 20 .82 seconds and the time saved we can say time saved would be equaling 26 .09 minus 20 .82.

02:57 So the time saved equals 5 .27 seconds.

03:08 This would be our final answer for the time saved by the racer...

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