/*! 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 11 The cheetah can reach a top spee... [FREE SOLUTION] | 91Ó°ÊÓ

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The cheetah can reach a top speed of \(114 \mathrm{~km} / \mathrm{h}\) \((71 \mathrm{mi} / \mathrm{h})\). While chasing its prey in a short sprint, a cheetah starts from rest and runs \(45 \mathrm{~m}\) in a straight line, reaching a final speed of \(72 \mathrm{~km} / \mathrm{h}\). (a) Determine the cheetah's average acceleration during the short sprint, and (b) find its displacement at \(t=3.5 \mathrm{~s}\).

Short Answer

Expert verified
The cheetah's average acceleration during the short sprint is \(4.44 \mathrm{~m/s}^2\) and its displacement at \(t = 3.5 \mathrm{~s}\) is \(27.33 \mathrm{~m}\).

Step by step solution

01

Convert units

Since the speed is given in \(\mathrm{km/h}\) and the distance in \(\mathrm{m}\), we need to convert the units to keep them consistent. We convert the final speed from \(\mathrm{km/h}\) to \(\mathrm{m/s}\) by multiplying \(72 \mathrm{~km/h}\) by \(1000 \mathrm{~m}/ \mathrm{km}\) and dividing by \(3600 \mathrm{~s}/ \mathrm{h}\), which results in \(20 \mathrm{~m/s}\).
02

Calculate average acceleration

We use the formula for acceleration, which is (final velocity - initial velocity) / time. We need to find the time. We know the cheetah runs \(45 \mathrm{~m}\) with a constant acceleration, so we use the kinematic equation \(d = vt + 0.5at^2\). Since the cheetah starts from rest, we can neglect the \(vt\) term. Solving for \(t\) we get \(t=\sqrt{(2d) / a}\). We substitute this back into the average acceleration formula, which gives us \(a = v_f^2 / (2d)\). Substituting \(v_f = 20 \mathrm{~m/s}\) and \(d = 45 \mathrm{~m}\) gives \(a = 4.44 \mathrm{~m/s^2}\).
03

Calculate displacement at t=3.5 s

We use the kinematic equation for the displacement, which is \(d = v_i*t + 0.5*a*t^2\). Since the cheetah starts from rest, the \(v_i*t\) term is eliminated. Substituting \(a = 4.44 \mathrm{~m/s^2}\) and \(t = 3.5 \mathrm{~s}\) gives \(d = 0.5 * 4.44 \mathrm{~m/s^2} * (3.5 \mathrm{~s})^2\), yielding \(d = 27.33 \mathrm{~m}\).

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Key Concepts

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

Average Acceleration
Average acceleration is a key concept in kinematics. It describes how the speed of an object changes over time. To calculate it, use the formula:\[a = \frac{v_f - v_i}{t}\]where:
  • \(a\) is average acceleration,
  • \(v_f\) is final velocity,
  • \(v_i\) is initial velocity, and
  • \(t\) is the time over which the change occurs.
In our cheetah example, it starts from rest (\(v_i = 0\)), so the equation simplifies to\[a = \frac{v_f}{t}\]Since the problem provides the final speed and distance, not the time, we rely on kinematic equations to find it. This exercise ultimately reveals the average acceleration of the cheetah during its sprint as \(4.44 \, \text{m/s}^2\). Understanding average acceleration helps us see how quickly the cheetah can speed up in real-life scenarios.
Unit Conversion
Unit conversion is crucial when dealing with physics problems, as it ensures we work with consistent units. In this exercise, you encounter speeds given in kilometers per hour (km/h), while the other measurements are in meters and seconds. This mismatch requires conversion to get accurate results.

Steps for Conversion

To convert speed from km/h to m/s, use the formula:\[v = \frac{v \, (\text{km/h}) \times 1000 \, (\text{m/km})}{3600 \, (\text{s/h})}\]By applying this, let's convert \(72 \, \text{km/h}\) to meters per second:
  • Multiply by \(1000\) to switch from km to m.
  • Divide by \(3600\) to switch from hours to seconds.
This conversion simplifies to:\[72 \, \text{km/h} \Rightarrow 20 \, \text{m/s}\]Remember to always convert your units before proceeding with calculations. This prevents mistakes and ensures the final answers are correct.
Kinematic Equations
Kinematic equations describe the motion of objects and are essential in solving problems involving acceleration, velocity, and displacement. These equations assume constant acceleration. In our example, we use one of these to find the average acceleration of the cheetah.

Relevant Equation

The kinematic equation used in this solution is:\[d = v_i \cdot t + 0.5 \, a \, t^2\]Here:
  • \(d\) is the displacement,
  • \(v_i\) is the initial velocity,
  • \(a\) is the acceleration, and
  • \(t\) is the time.
Since the cheetah starts at rest, \(v_i = 0\), simplifying the equation to:\[d = 0.5 \, a \, t^2\]Using this, we discovered the cheetah's acceleration and calculated the displacement after \(3.5\) seconds. Practicing with such equations will sharpen your ability to analyze complex motion scenarios effectively.

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Most popular questions from this chapter

Two students are on a balcony a distance \(h\) above the street. One student throws a ball vertically downward at a speed \(v_{0} ;\) at the same time, the other student throws a ball vertically upward at the same speed. Answer the following symbolically in terms of \(v_{0}, g\), \(h\), and \(t\). (a) Write the kinematic equation for the \(y\)-coordinate of each ball. (b) Set the equations found in part (a) equal to height 0 and solve each for \(t\) symbolically using the quadratic formula. What is the difference in the two balls' time in the air? (c) Use the time-independent kinematics equation to find the velocity of each ball as it strikes the ground. (d) How far apart are the balls at a time \(t\) after they are released and before they strike the ground?

A mountain climber stands at the top of a \(50.0-\mathrm{m}\) cliff that overhangs a calm pool of water. She throws two stones vertically downward \(1.00 \mathrm{~s}\) apart and observes that they cause a single splash. The first stone had an initial velocity of \(-2.00 \mathrm{~m} / \mathrm{s}\). (a) How long after release of the first stone did the two stones hit the water? (b) What initial velocity must the second stone have had, given that they hit the water simultaneously? (c) What was the velocity of each stone at the instant it hit the water?

One athlete in a race running on a long, straight track with a constant speed \(v_{1}\) is a distance \(d\) behind a second athlete running with a constant speed \(v_{2}\). (a) Under what circumstances is the first athlete able to overtake the second athlete? (b) Find the time \(t\) it takes the first athlete to overtake the second athlete, in terms of \(d, v_{1}\), and \(v_{2}\). (c) At what minimum distance \(d_{2}\) from the leading athlete must the finish line be located so that the trailing athlete can at least tie for first place? Express \(d_{2}\) in terms of \(d, v_{1}\), and \(v_{2}\) by using the result of part (b).

A stuntman sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is \(10.0 \mathrm{~m} / \mathrm{s}\), and the man is initially \(3.00 \mathrm{~m}\) above the level of the saddle. (a) What must be the horizontal distance between the saddle and the limb when the man makes his move? (b) How long is he in the air?

A tortoise can run with a speed of \(0.10 \mathrm{~m} / \mathrm{s}\), and a hare can run 20 times as fast. In a race, they both start at the same time, but the hare stops to rest for \(2.0\) minutes. The tortoise wins by a shell \((20 \mathrm{~cm})\). (a) How long does the race take? (b) What is the length of the race?

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