/*! 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

A steam catapult launches a jet aircraft from the aircraft carrier John C. Stennis, giving it a speed of \(175 \mathrm{mi} / \mathrm{h}\) in \(2.50 \mathrm{~s}\). (a) Find the average acceleration of the plane. (b) Assuming the acceleration is constant, find the distance the plane moves.

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of \(12 \mathrm{~m} / \mathrm{s}\), skates by with the puck. After \(3.0 \mathrm{~s}\), the first player makes up his mind to chase his opponent. If he accelerates uniformly at \(4.0 \mathrm{~m} / \mathrm{s}^{2}\), (a) how long does it take him to catch his opponent, and (b) how far has he traveled in that time? (Assume the player with the puck remains in motion at constant speed.)

Two boats start together and race across a \(60-\mathrm{km}\)-wide lake and back. Boat A goes across at \(60 \mathrm{~km} / \mathrm{h}\) and returns at \(60 \mathrm{~km} / \mathrm{h}\). Boat B goes across at \(30 \mathrm{~km} / \mathrm{h}\), and its crew, realizing how far behind it is getting, returns at \(90 \mathrm{~km} / \mathrm{h}\). Turnaround times are negligible, and the boat that completes the round trip first wins. (a) Which boat wins and by how much? (Or is it a tie?) (b) What is the average velocity of the winning boat?

A car starts from rest and travels for \(5.0 \mathrm{~s}\) with a uniform acceleration of \(+1.5 \mathrm{~m} / \mathrm{s}^{2}\). The driver then applies the brakes, causing a uniform acceleration of \(-2.0 \mathrm{~m} / \mathrm{s}^{2}\). If the brakes are applied for \(3.0 \mathrm{~s}\), (a) how fast is the car going at the end of the braking period, and (b) how far has the car gone?

A race car moves such that its position fits the relationship $$ x=(5.0 \mathrm{~m} / \mathrm{s}) t+\left(0.75 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} $$ where \(x\) is measured in meters and \(t\) in seconds. (a) Plot a graph of the car's position versus time. (b) Determine the instantaneous velocity of the car at \(t=4.0 \mathrm{~s}\), using time intervals of \(0.40 \mathrm{~s}, 0.20 \mathrm{~s}\), and \(0.10 \mathrm{~s}\). (c) Compare the average velocity during the first \(4.0 \mathrm{~s}\) with the results of part (b).

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