The leader of a bicycle race is traveling with a constant...

The leader of a bicycle race is traveling with a constant velocity of $+11.10 \mathrm{m} / \mathrm{s}$ and is 10.0 $\mathrm{m}$ ahead of the second-place cyclist. The second-place cyclist has a velocity of $+9.50 \mathrm{m} / \mathrm{s}$ and an acceleration of $+1.20 \mathrm{m} / \mathrm{s}^{2}$ . How much time elapses before he catches the leader?

Key Concepts

Quadratic Equation Solving

In problems involving constant acceleration, solving for time often leads to a quadratic equation. The ability to solve such equations, either by factoring or by using the quadratic formula, is critical for finding the time at which two moving objects have the same displacement. This mathematical technique is a common tool in kinematics problems.

Relative Motion Analysis

This key concept involves comparing the motion of two objects to determine when they will meet or catch up with one another. By setting the displacement equations of two objects equal to each other, one can solve for the time at which their positions coincide. This analysis is particularly useful when one object is traveling at a constant velocity and another is accelerating.

Kinematics Equations for Constant Velocity

This concept involves the understanding that when an object moves at a constant velocity, its displacement increases linearly with time. The relationship is described by the simple equation: displacement equals velocity multiplied by time. This principle is fundamental for modeling the motion of objects that do not accelerate.

Kinematics Equations for Constant Acceleration

When an object experiences constant acceleration, its displacement as a function of time is given by a quadratic relationship. The standard equation used is: displacement equals initial velocity times time plus one-half times acceleration times the square of time. This equation allows one to determine how the distance traveled by an accelerating object changes over time.

Transcript

00:01 Okay, so in this problem, we're working with two bicyclists that are in a race.

00:06 Our first one is the leader, and the second one is playing catch -up.

00:10 Now, we know that the leader is traveling at a rate of 11 .1 meters per second, and that this is a constant velocity, meaning the acceleration is zero meters per second.

00:22 The racer who's playing catch -up is starting at a slower rate of 9 .5 meters per second, and his acceleration is 1 .2 meters per second squared.

00:34 Now, the question's asking us, how long will it take for the person playing catchup to catch the leader? so the time is going to be the same value, but the leader is 10 meters in front of the person playing catch up.

00:45 This means that however far the leader travels, the person playing ketchup has to travel that far, and they have to travel 10 additional meters because they're trying to make up that distance between them.

00:57 So for both of these equations, we do not know anything about the final velocity.

01:01 So the kinematic equation we're going to be using to solve them is that last one there because it does not have the final velocity in it.

01:07 So let's do the leader's equation first.

01:10 His x value is x equals v0 t, so 11 .1 multiplied by time, plus one -half a t squared.

01:20 Now the nice thing here is there's no acceleration for the leader.

01:23 So this is actually the end of that equation.

01:25 And this gives us a nice term for x that we can use to substantiated.

01:29 In to the person playing catch up.

01:32 So switching over to the second person's equation, going with the same one, the x equals v .0 t plus one half a t squared.

01:38 Instead of x, i have x plus 10 equals the initial velocity multiplied by time plus one half a t squared, which gives us 0 .6t squared...

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