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Chapter 2: Problem 79

A cheetah is hunting. Its prey runs for \(3.0 \mathrm{s}\) at a constant velocity of \(+9.0 \mathrm{m} / \mathrm{s}\). Starting from rest, what constant acceleration must the cheetah maintain in order to run the same distance as its prey runs in the same time?

Short Answer

Expert verified
The cheetah must maintain an acceleration of \(6.0 \text{ m/s}^2\).

Step by step solution

01

Calculate Distance Run by the Prey

To find out how far the prey travels, use the formula for distance, \( d = vt \), where \( v \) is the velocity and \( t \) is the time. Substitute the given values \( v = 9.0 \) m/s and \( t = 3.0 \) s.\[ d = 9.0 \, \text{m/s} \times 3.0 \, \text{s} = 27.0 \, \text{m} \]
02

Set Up the Equation for the Cheetah's Motion

The cheetah starts from rest, so its initial velocity \( u = 0 \). Use the equation \( d = ut + \frac{1}{2}at^2 \) to find the acceleration \( a \). Substitute \( d = 27.0 \) m, \( u = 0 \), and \( t = 3.0 \) s.
03

Solve for the Cheetah's Acceleration

Substitute \( u = 0 \) into the equation \( d = ut + \frac{1}{2}at^2 \), it simplifies to \( d = \frac{1}{2}at^2 \). Solve for \( a \):\[ 27.0 = \frac{1}{2}a(3.0)^2 \]\[ 27.0 = \frac{1}{2}a \times 9.0 \]\[ 27.0 = 4.5a \]Divide both sides by 4.5 to find \( a \):\[ a = \frac{27.0}{4.5} = 6.0 \, \text{m/s}^2 \]

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Key Concepts

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

Constant Velocity
Constant velocity means that an object moves at the same speed and in the same direction over time. In the original exercise, the prey runs with a constant velocity of \(9.0 \text{ m/s}\). This is a typical scenario in kinematics where the speed of the object does not change.

Constant velocity is interesting because:
  • The speed stays the same, so the distance covered over time is predictable.
  • No acceleration is involved, meaning there's no change in speed or direction.
  • The distance can be calculated easily using the formula \(d = vt\), where \(d\) is distance, \(v\) is velocity, and \(t\) is time.
Understanding constant velocity is crucial because it simplifies the analysis of moving objects when acceleration is not a factor.
Acceleration
Acceleration describes the change in velocity over time. If the speed of an object increases or decreases, or if the object's direction changes, it is experiencing acceleration. In the exercise, the cheetah has to accelerate from rest to reach the distance the prey covers.

Here鈥檚 what acceleration involves:
  • It is calculated as \(a = \frac{\Delta v}{\Delta t}\), where \(\Delta v\) is the change in velocity, and \(\Delta t\) is the change in time.
  • Positive acceleration indicates an increase in velocity, while negative (deceleration) means a decrease.
  • In our scenario, the cheetah needs a constant acceleration of \(6.0 \text{ m/s}^2\) to match the prey's distance in the same timeframe.
Understanding acceleration helps to predict how quickly an object can change its speed, which is essential in many real-world applications.
Distance Formula
The distance formula \(d = vt\) is used to calculate the distance an object travels with a constant velocity. It becomes more complex when acceleration is involved.

In scenarios involving acceleration, we use another formula:
  • \(d = ut + \frac{1}{2}at^2\), where \(d\) is distance, \(u\) is initial velocity, \(a\) is acceleration, and \(t\) is time.

In the solution, the prey covers \(27.0 \text{ m}\) using the simpler constant velocity formula. For the cheetah, which starts from rest with \(u=0\), the distance formula adapts to find how far it needs to travel under constant acceleration.

These formulas are foundational in kinematics, helping to understand and calculate motion under different conditions.

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Most popular questions from this chapter

A VW Beetle goes from 0 to \(60.0 \mathrm{mi} / \mathrm{h}\) with an acceleration of \(+2.35 \mathrm{m} / \mathrm{s}^{2}\) (a) How much time does it take for the Beetle to reach this speed? (b) A top-fuel dragster can go from 0 to \(60.0 \mathrm{mi} / \mathrm{h}\) in \(0.600 \mathrm{s}\). Find the acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ) of the dragster.

A cement block accidentally falls from rest from the ledge of a \(53.0-\mathrm{m}\) -high building. When the block is \(14.0 \mathrm{m}\) above the ground, a man, \(2.00 \mathrm{m}\) tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?

For each of the three pairs of positions listed in the following table, determine the magnitude and direction (positive or negative) of the displacement. $$ \begin{array}{lcc} \hline & \text { Initial position } x_{0} & \text { Final position } x \\ \hline \text { (a) } & +2.0 \mathrm{m} & +6.0 \mathrm{m} \\ \text { (b) } & +6.0 \mathrm{m} & +2.0 \mathrm{m} \\ \text { (c) } & -3.0 \mathrm{m} & +7.0 \mathrm{m} \\ \hline \end{array} $$

A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the first part, she rides for 22 minutes at an average speed of \(7.2 \mathrm{m} / \mathrm{s}\). During the second part, she rides for 36 minutes at an average speed of \(5.1 \mathrm{m} / \mathrm{s}\). Finally, during the third part, she rides for 8.0 minutes at an average speed of \(13 \mathrm{m} / \mathrm{s}\). (a) How far has the bicyclist traveled during the entire trip? (b) What is her average velocity for the trip?

A hot-air balloon is rising upward with a constant speed of \(2.50 \mathrm{m} / \mathrm{s}\). When the balloon is \(3.00 \mathrm{m}\) above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?
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