/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 The determined coyote is out onc... [FREE SOLUTION] | 91Ó°ÊÓ

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The determined coyote is out once more in pursuit of 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. P4.65). The coyote starts at rest 70.0 \(\mathrm{m}\) from the brink 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 in order to reach the cliff before the coyote. At the edge of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. His skates remain horizontal and continue to operate while he is in flight, so that the coyote’s acceleration while in the air is \((15.0 \hat{\mathrm{i}}-9.80 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2} .\) (b) If the cliff is 100 \(\mathrm{m}\) above the flat floor of a canyon, determine where the coyote lands in the canyon. (c) Determine the components of the coyote's impact velocity.

Short Answer

Expert verified
The roadrunner's minimum speed is obtained by dividing the distance by the time for the coyote to reach the cliff. The horizontal distance the coyote travels in the air is found using the horizontal motion equation with the air time. The impact velocity components are the horizontal velocity at the end of the air time, and the vertical velocity obtained from the vertical motion equation.

Step by step solution

01

Determine Time for Coyote to Reach the Cliff

To find the time it takes for the coyote to reach the brink of the cliff, use the kinematic equation for constant acceleration: \(v = u + at\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. Since the coyote starts at rest, \(u = 0\). The coyote's acceleration is 15.0 m/s^2 and he must cover a distance of 70.0 m. We use the equation \(s = ut + \frac{1}{2}at^2\), where \(s\) is the distance. Solving for \(t\) gives \(t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{2 \times 70.0}{15.0}}\).
02

Calculate the Roadrunner's Minimum Speed

The roadrunner must reach the cliff in the same time or less than it takes the coyote. Using the time calculated in Step 1, we set the distance that the roadrunner must travel (70.0 m) equal to the product of the roadrunner’s speed and the time obtained. Let \(v_r\) be the roadrunner's constant speed and \(t\) be the time calculated before. Then \(v_r = \frac{s}{t}\).
03

Calculate the Coyote's Air Time

To find the time the coyote is in the air, use the vertical motion equation \(y = v_{0y}t + \frac{1}{2}a_yt^2\). The initial vertical velocity \(v_{0y} = 0\) since he moves horizontally off the cliff. The vertical acceleration \(a_y = -9.80 \mathrm{m/s^2}\) due to gravity. The vertical distance \(y = -100\) m since the coyote falls to the canyon floor. Solve the equation \(0 = -100 - \frac{1}{2}(9.80)t^2\) to find the time \(t\) in the air.
04

Calculate the Horizontal Distance Coyote Travels While in the Air

The horizontal acceleration \(a_x = 15.0 \mathrm{m/s^2}\), and since the coyote keeps accelerating horizontally while falling, we use \(x = v_{0x}t + \frac{1}{2}a_x t^2\). The initial horizontal velocity \(v_{0x}\) will be the velocity reached at the edge of the cliff, which is \(v = at\), with \(t\) being the time to the edge of the cliff calculated previously. Use the air time from Step 3 to calculate the total horizontal distance \(x\).
05

Determine the Components of the Coyote's Impact Velocity

The coyote's impact velocity will have both horizontal and vertical components. The horizontal component \(v_{ix}\) will be equal to the horizontal velocity at the end of the air time: \(v_{ix} = v_{0x} + a_x t_{air}\). The vertical component \(v_{iy}\) can be calculated using the equation \(v_{iy} = v_{0y} + a_y t_{air}\). The final impact velocity components are obtained by substituting \(v_{0x}\), \(a_x\), \(v_{0y}\), \(a_y\), and the air time \(t_{air}\) obtained in the previous steps.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Acceleration
When we talk about constant acceleration, we're discussing a situation where an object's velocity is changing at a consistent rate over time. This is a fundamental concept in kinematics, the branch of physics that deals with motion without considering the forces that cause it.

In the given exercise, the coyote's Acme jet-powered roller skates provide a steady acceleration, meaning regardless of other factors like friction or air resistance (which are typically ignored in basic kinematics problems), the coyote's velocity will increase by the same amount every second. Understanding this concept is crucial for analyzing his motion using the kinematics equations, which are specifically designed for constant acceleration situations.

The equation of motion for a linearly accelerating object is \[ s = ut + \frac{1}{2}at^2 \. \] Here, \(s\) represents the displacement, \(u\) is initial velocity, \(a\) is the constant acceleration, and \(t\) is the time. Since the initial velocity \(u\) is zero for the coyote starting at rest, the equation becomes simpler and extremely useful to determine the time needed for him to reach the cliff.
Kinematics in Two Dimensions
Kinematics in two dimensions, also known as projectile motion, comes into play when an object moves in a plane, involving both vertical and horizontal components of motion. For our roadrunner-chasing coyote, this becomes relevant the moment he shoots off the cliff edge.

In two-dimensional kinematics, we separately analyze the horizontal and vertical motions since they are independent of each other under constant acceleration, such as gravity. For the coyote, we ignore air resistance, meaning the only vertical force acting on him is gravity, causing a constant downward acceleration. Horizontally, he's still powered by his skates, leading to a constant forward acceleration.

To find the landing spot in the canyon, we use the equations \(x = v_{0x}t + \frac{1}{2}a_x t^2\) for horizontal motion and \(y = v_{0y}t + \frac{1}{2}a_y t^2\) for vertical motion (with \(v_{0y} = 0\) since there is no initial vertical velocity). It's this separation of dimensions that makes predicting the coyote's landing spot possible, despite the complexities of his motion.
Projectile Motion
Projectile motion is a form of motion experienced by an object that is thrown near the Earth's surface and moves along a curved path under the action of gravity only. This kind of motion is two-dimensional since it includes both a horizontal and a vertical component.

In our textbook problem, the coyote on roller skates becomes a projectile the moment he zooms off the edge of the cliff. While he unintentionally subjects himself to projectile motion, we can determine his trajectory and landing point using kinematics.

The coyote's situation is even more special because he continues to accelerate horizontally due to his roller skates even while in the air, which adds a twist to the usual projectile motion scenario. Usually, projectiles experience zero horizontal acceleration unless propelled, as in the case of rockets or, in our comedic setting, a coyote on powered skates. For his vertical motion, the only acceleration is due to gravity, which is consistently acting downwards at \(9.80 \mathrm{m/s^2}\).

To calculate his landing point, you would need to determine how long he's in the air using the equation \(y = v_{0y}t + \frac{1}{2}a_y t^2\) and then calculate the horizontal distance traveled using \(x = v_{0x}t + \frac{1}{2}a_x t^2\), both equations showcasing the quintessential characteristics of projectile motion.

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