The determined coyote is out once more to try to capture the...

The determined coyote is out once more to try to capture the elusive roadrunner. The coyote wears a pair of Acme jet-powered roller skates, which provide a constant horizontal acceleration of $15.0 \mathrm{~m} / \mathrm{s}^{2}$ (Fig. $\left.\mathrm{P} 4.66\right)$. The coyote starts off at rest $70.0 \mathrm{~m}$ from the edge of a cliff at the instant the roadrunner zips past him in the direction of the cliff. (a) If the roadrunner moves with constant speed, determine the minimum speed he must have to reach the cliff before the coyote. At the brink of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. (b) If the cliff is $100 \mathrm{~m}$ above the floor of a canyon, determine where the coyote lands in the canyon (assume his skates remain horizontal and continue to operate when he is in "flight" ). (c) Determine the components of the coyote's impact velocity.

Key Concepts

Constant Acceleration

This concept involves the motion of an object when it experiences a fixed, unchanging acceleration. In problems of constant acceleration, equations of motion (such as displacement, velocity, and time relationships) can be used to predict the object's future state. These equations help determine positions, speeds, and timings based on the initial conditions and the constant acceleration provided by forces or machinery.

Uniform Linear Motion

Uniform linear motion describes the movement of an object at a constant speed in a straight line, meaning there is no acceleration acting on the object. This concept is essential for determining positions based solely on the product of speed and time, and it is often contrasted with accelerated motion in problems involving pursuit or race scenarios.

Projectile Motion

Projectile motion occurs when an object is launched into the air and moves under the influence of gravity, following a curved, parabolic path. In such cases, the horizontal and vertical components of motion are considered separately; the horizontal motion typically remains uniform (if air resistance is neglected), while the vertical motion is uniformly accelerated due to gravity.

Separation of Motion Components

This principle states that the horizontal and vertical motions of a projectile are independent of each other. Horizontal motion can be analyzed using uniform motion equations while vertical motion is solved using constant acceleration due to gravity. This separation is essential for solving problems where an object moves both horizontally (with or without additional acceleration) and vertically (subject to gravitational acceleration).

Transcript

00:02 Okay, so for the first part of this question, it wants us to find the minimum speed the roadrunner must have to be at before the coyote reaches him at the edge of the cliff.

00:15 So we have a few things given to us over here.

00:19 We have the constant acceleration of the coyote at 15.

00:25 The final distance, which is 70, is how far away the cliff is.

00:29 We have his initial velocity which is zero and the initial position which is zero.

00:33 So if we use our first kinematic equation here, up here on the top right, we can use that to determine the time it takes for the coyote to get to the edge of the cliff, which is what we'll need in order to, which is what we'll need in order to find how long it takes for the roadrunner to reach him.

01:10 So we'll do that right now.

01:14 So if we plug some things in, we can determine what the time is.

01:19 So we have x is 70, x of 0 plus 0 times time, which is 0, plus 1 .5 times 15 times times squared.

01:36 So if we rearrange this equation, we can get that 15 divided by 2 is 7 .5.

01:44 And we take that away from 70.

01:48 So we get 62 .5 equals time squared.

01:55 So time is the square root of that, which is about 7 .9.

02:11 And that'll be the time it takes, 7 .9 seconds for the coyote to reach the edge of the cliff.

02:18 And so now we can use the normal velocity equation.

02:23 Velocity equals distance over time.

02:25 To determine the speed the rudder has to be at in order to reach the cliff before the coyote does or at least at the same time so we'll use the distance which is 70 over the time which is 7 .9 which gets us 8 .85 about meters per second so that is our answer for part a and i'll box that for you guys okay now let's move on to part b.

03:08 So in part b, it says at the brink of the cliff, the roadrunner escapes by making a sudden turn while the coyote turns straight ahead.

03:18 If the cliff is 100 meters above the floor of a canyon, determine where the coyote lands in the canyon, assuming his skates remain horizontal and continuing to operate when he's in flight.

03:29 Okay, so this is going to be a parabolic gravity equation.

03:34 So we need to determine the time it takes for the coyote to reach the ground and then we'll use that time to get how far he goes.

03:48 So if we do that, we'll need to do the y direction of the coyote.

03:58 So he starts at 100 meters above the ground.

04:02 So we'll say y is equal to 100 meters and then we'll say that's his initial so y initials 100 why is and then the final y is zero that's when he hits the ground so we know the acceleration the acceleration of gravity is 9 .8 and so because the coyote is falling it's in the negative direction negative 9 .8 meters per second squared so now we can use we can use this same equation we used at the up here to find the time it takes for the coyote to reach the ground.

04:50 Because there's no lift happening.

04:52 So the coyote's just going to fall in the y direction as if he was falling off the cliff without going anywhere.

04:58 So we have y equals y0 plus velocity in the y direction initial times time plus one half times gravity times times squared.

05:17 So we know the the final y is 0, the initial y is 100, the initial y velocity is 0 times time which is 0, plus 1 1⁄2 times 9 .8 times times squared.

05:36 So we can do some algebra here to get that negative 9 .8 divided by 2 is equal to negative 4 .9 times 100 is negative 490 and we pop that to the other side.

05:54 Of the equation, oh sorry.

06:01 So we get negative 100 on this side of the equation, and then you get negative 9 .8 times 1 .5, which is negative 4 .9 t squared, and then we do 100 divided by 4 .9, which is 20 .4.

06:25 So t squared equals about 20 .4 and t is the square root of that, which is about 4 .5 seconds.

06:42 So that's the time it takes for the coyote to hit the ground.

06:45 So now we can take that time to determine how far the coyote goes with this constant acceleration in the x direction.

06:55 So what we're going to do is we're going to use the velocity we got from this past equation.

07:06 Oh well, we need to find his final velocity, and then we need to use that final velocity at the top of the cliff in the x direction to determine how far he goes or how far he'd keep going.

07:22 So we know that the coyote took 7 .9 seconds to get to the edge of the cliff.

07:29 So if we're going to find the final velocity of the coyote, we'll do the final equals the initial plus a times time.

07:40 So we know the initial is zero...

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