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Chapter 2: Problem 48

A chectah can accelerate from rest to \(25.0 \mathrm{m} / \mathrm{s}\) in \(6.22 \mathrm{s}\). Assuming constant acceleration, (a) how far has the chectah run in this time? (b) After sprinting for just 3.11 s, is the chectah's specd \(12.5 \mathrm{m} / \mathrm{s},\) more than \(12.5 \mathrm{m} / \mathrm{s},\) or less than \(12.5 \mathrm{m} / \mathrm{s}\) ? Explain. \((\mathrm{c})\) What is the cheetah's average speed for the first 3.11 s of its sprint? For the second 3.11 s of its sprint? (d) Calculate the distance covered by the cheetah in the first 3.11 s and the second 3.11 s.

Short Answer

Expert verified
The cheetah runs 77.56 m in 6.22 s. After 3.11 s, it is moving at 12.5 m/s. The cheetah's average speed is 6.25 m/s for the first 3.11 s and 18.75 m/s for the second.

Step by step solution

01

Understanding the Equation for Constant Acceleration

The cheetah's acceleration is constant, so we use the equation for displacement: \( s = ut + \frac{1}{2}at^2 \), where \( s \) is the displacement, \( u \) is the initial velocity (0 in this case since it starts from rest), \( a \) is the acceleration, and \( t \) is the time.
02

Calculate the Acceleration

First, we use the formula for acceleration: \( a = \frac{v - u}{t} \). Given that \( u = 0 \), \( v = 25.0 \text{ m/s} \), and \( t = 6.22 \text{ s} \), the acceleration \( a \) is \( \frac{25.0}{6.22} = 4.02 \text{ m/s}^2 \).
03

Calculate the Total Distance in 6.22 seconds

Plug the acceleration value \( a = 4.02 \text{ m/s}^2 \) into the displacement equation: \( s = 0 + \frac{1}{2} \times 4.02 \times (6.22)^2 \). Calculate \( s \) to find the distance: \( s = 77.56 \text{ m} \).
04

Determine Speed at 3.11 Seconds

To find the cheetah's speed at 3.11 seconds, use \( v = u + at \). So \( v = 0 + 4.02 \times 3.11 = 12.50 \text{ m/s} \). The speed is exactly 12.5 m/s.
05

Find Average Speed for First 3.11 Seconds

The initial speed is 0 and final speed at 3.11s is 12.5 m/s, so average speed is \( \frac{0 + 12.5}{2} = 6.25 \text{ m/s} \).
06

Find Average Speed for Second 3.11 Seconds

The speed at the start of the second half is 12.5 m/s and at the end of 6.22 seconds is 25 m/s. So the average speed is \( \frac{12.5 + 25.0}{2} = 18.75 \text{ m/s} \).
07

Calculate Distance for First 3.11 Seconds

Using the distance equation for the first 3.11 seconds: \( s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}\times 4.02 \times (3.11)^2 \). This gives \( s = 19.39 \text{ m} \).
08

Calculate Distance for Second 3.11 Seconds

Distance in the second 3.11 seconds is the total distance minus the first 3.11 seconds: \( s_{total} = 77.56 \text{ m} \), first 3.11 s = 19.39 m, thus \( s_{second} = 77.56 - 19.39 = 58.17 \text{ m} \).

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Key Concepts

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

Kinematics
Kinematics is the branch of mechanics that primarily deals with the concepts related to motion. When analyzing the movement of objects, specifically those in a straight line with constant acceleration, kinematics provides us with valuable tools and equations.
In our exercise, we deal with a cheetah undergoing constant acceleration. To understand this motion, we focus on three main variables: initial velocity, final velocity, and time. Initial velocity is the speed at which the cheetah starts, which in this case is zero since it starts from rest. Final velocity is the speed attained after a given period. Time is the interval over which this change occurs.
The fundamental kinematics equations help us calculate unknown values like displacement, speed, and acceleration. These calculations are essential in predicting the cheetah's behavior over time, given its initial conditions. With constant acceleration, one of the key equations is the displacement equation which reveals more about how much ground the cheetah would cover in a given time period.
Displacement Equation
The displacement equation is vital in analyzing the distance covered by an object under constant acceleration. It allows us to calculate how far an object travels in a specific timeframe. The equation is given by:
  • \( s = ut + \frac{1}{2}at^2 \)
This equation highlights several factors:
  • Initial Velocity \( (u) \): This is the starting speed of the object, zero when the object starts from rest.
  • Time \( (t) \): Represents how long the object has been moving.
  • Acceleration \( (a) \): The rate at which the velocity of the object changes, calculated using the formula \( a = \frac{v-u}{t} \).
  • Displacement \( (s) \): The total distance travelled by the object.
Understanding each variable helps in accurately determining the distance travelled. By substituting known values into the formula, like in our cheetah example, you can find the total displacement after a given period. With constant acceleration, this equation simplifies calculations and reliably predicts the object's future position.
Average Speed
Average speed gives insight into the overall speed of an object over a period of time. It is calculated as the total displacement divided by the time taken to achieve that displacement.
When considering average speed over intervals with constant acceleration, the average speed is the mean of the initial and final speeds within that interval.
This can be represented by:
  • \( \text{Average Speed} = \frac{u + v}{2} \)
In our cheetah scenario, understanding average speed is important to split the running time into two equal segments (first and second halves of the sprint), allowing us to see how the cheetah's speed changes over each part.
This concept helps create a more comprehensive view of the cheetah's motion, showing not just speed at an instant but the overall efficiency of its run over time. Average speed can differ in shorter intervals even under constant acceleration, as seen when calculating the average for the first and second 3.11 seconds of the cheetah's sprint. This gives us a complete insight into how the speed evolves over time.

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