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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 needs an acceleration of 6.0 m/s虏.

Step by step solution

01

Calculate the Distance Covered by the Prey

The prey travels with a constant velocity of \(9.0 \, \mathrm{m/s}\) for \(3.0 \, \mathrm{s}\). The distance covered \(d\) can be calculated using the formula: \[ d = v \times t \]Substituting the given values, \[ d = 9.0 \, \mathrm{m/s} \times 3.0 \, \mathrm{s} = 27.0 \, \mathrm{m} \]The prey covers a distance of \(27.0 \, \mathrm{m}\).
02

Relate the Cheetah's Motion to Distance

The cheetah must run exactly \(27.0 \, \mathrm{m}\) in the same time of \(3.0 \, \mathrm{s}\). Since the cheetah starts from rest, it uses the constant acceleration formula: \[ d = \frac{1}{2} a t^2 \]where \(d\) is the distance to be covered, \(a\) is the acceleration, and \(t\) is the time.
03

Solve for the Cheetah's Required Acceleration

We know that \(d = 27.0 \, \mathrm{m}\) and \(t = 3.0 \, \mathrm{s}\). Substitute these values into the formula: \[ 27.0 \, \mathrm{m} = \frac{1}{2} a (3.0 \, \mathrm{s})^2 \]\[ 27.0 \, \mathrm{m} = \frac{1}{2} a \times 9.0 \, \mathrm{s}^2 \]\[ 27.0 \, \mathrm{m} = 4.5 \, a \, \mathrm{s}^2 \]Divide both sides by \(4.5 \, \mathrm{s}^2\) to solve for \(a\): \[ a = \frac{27.0 \, \mathrm{m}}{4.5 \, \mathrm{s}^2} = 6.0 \, \mathrm{m/s^2} \]The cheetah must maintain a constant acceleration of \(6.0 \, \mathrm{m/s^2}\).

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Key Concepts

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

Constant Velocity
Constant velocity occurs when an object moves at a steady rate without changing its speed or direction. In the exercise above, the prey runs with a constant velocity of 9.0 m/s, which means it travels the same distance every second. Understanding constant velocity is crucial for solving problems related to uniform motion, as it simplifies calculations by allowing the use of basic formulas.
To calculate the distance covered under constant velocity, you can use the simple formula:
  • \( d = v \times t \)
Here, \(d\) is the distance, \(v\) is the velocity, and \(t\) is the time. This equation helps quickly determine distances when speed doesn't change.
Constant Acceleration
Constant acceleration describes a scenario where an object's velocity changes by the same amount each second. In our problem, the cheetah starts from rest and aims to match the distance of the prey by accelerating steadily.
Constant acceleration can be described by the following formula:
  • \( a = \frac{v_{f} - v_{i}}{t} \)
Where \(a\) is the acceleration, \(v_{f}\) is the final velocity, \(v_{i}\) is the initial velocity, and \(t\) is the time duration. For the cheetah, though it starts at rest, it's more useful to use the distance-related equation here since we want to calculate how much acceleration it needs to travel 27 meters in 3 seconds.
Distance Calculation
Calculating distance involves understanding the relationship between speed, time, and acceleration鈥攄epending on the type of motion. For constant velocity, like the prey's motion, distance \(d\) is given by:
  • \( d = v \times t \)
In contrast, for an object with constant acceleration starting from rest, like the cheetah, the equation becomes:
  • \( d = \frac{1}{2} a t^2 \)
This formula accounts for the increasing speed as the cheetah accelerates. By rearranging it, you can solve for acceleration or the other variables, depending on the known quantities and what you need to find.
Motion Equations
Kinematic equations, also known as the equations of motion, are key to solving different types of motion problems. They describe an object's velocity, acceleration, distance, and time without needing to know the forces involved.
For constant velocity, motion is simple, and you only need one equation:
  • \( d = v \times t \)
For constant acceleration, other equations come into play, such as:
  • \( v = v_{i} + a \times t \)
  • \( d = v_{i} \times t + \frac{1}{2} a t^2 \)
These allow us to determine unknowns by rearranging them based on the information provided. Applying these principles helps solve many real-world problems involving motion, like that of a fast-moving cheetah accelerating to catch its prey.

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Most popular questions from this chapter

Over a time interval of 2.16 years, the velocity of a planet orbiting a distant star reverses direction, changing from \(+20.9 \mathrm{km} / \mathrm{s}\) to \(-18.5 \mathrm{km} / \mathrm{s} .\) Find (a) the total change in the planet's velocity (in \(\mathrm{m} / \mathrm{s}\) ) and (b) its average acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.

A Deadly Virus. On your flight out of South America your cargo plane emergency-lands on a deserted island somewhere in the South Pacific. Because it was a beach landing, the pilot skids the plane on its belly rather than lowering the landing gear. The landing is rough: an initial vertical drop followed by an abrupt horizontal deceleration. When the dust settles, the pilot and your scientist colleagues are all okay, but you know there might be a big problem. Your team is transporting samples of a deadly virus from a recent breakout in a small village. The samples were packed in a cryogenically cooled container that has a special seal that may leak if it exceeds an acceleration greater than 10 g. The container has a mechanical accelerometer gauge that measures the maximum vertical deceleration in case the container is dropped. It reads a value of \(a=7.30\) g. However, you are still worried: there was another component of the acceleration. You get out of the plane and measure the length of the skid the plane made in the sand as it landed: 154 ft. The pilot claims the plane's horizontal speed on impact was \(160 \mathrm{m} \cdot \mathrm{p} . \mathrm{h} .\) Assuming the worst case scenario (i.e., that the vertical and horizontal accelerations occurred simultaneously), and that the horizontal deceleration was uniform, is it likely that the seal on the virus container is compromised? Explain.

Two arrows are shot vertically upward. The second arrow is shot after the first one, but while the first is still on its way up. The initial speeds are such that both arrows reach their maximum heights at the same instant, although these heights are different. Suppose that the initial speed of the first arrow is \(25.0 \mathrm{m} / \mathrm{s}\) and that the second arrow is fired \(1.20 \mathrm{s}\) after the first. Determine the initial speed of the second arrow.

A ball is thrown vertically upward, which is the positive direction. A little later it returns to its point of release. The ball is in the air for a total time of 8.0 s. What is its initial velocity? Neglect air resistance.

The initial velocity and acceleration of four moving objects at a given instant in time are given in the following table. Determine the final speed of each of the objects, assuming that the time elapsed since \(t=0\) s is 2.0 s. $$ \begin{array}{lcc} & \text { Initial velocity } v_{0} & \text { Acceleration } a \\\\\hline \text { (a) } & +12 \mathrm{m} / \mathrm{s} & +3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (b) } & +12 \mathrm{m} / \mathrm{s} & -3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (c) } & -12 \mathrm{m} / \mathrm{s} & +3.0 \mathrm{m} / \mathrm{s}^{2} \\\\\hline \text { (d) } & -12 \mathrm{m} / \mathrm{s} & -3.0 \mathrm{m} / \mathrm{s}^{2} \\ \hline\end{array}$$
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