/*! 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 97 A runner approaches the finish l... [FREE SOLUTION] | 91Ó°ÊÓ

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A runner approaches the finish line and is \(75 \mathrm{m}\) away; her average speed at this position is \(8 \mathrm{m} / \mathrm{s} .\) She decelerates at this point at \(0.5 \mathrm{m} / \mathrm{s}^{2} .\) How long does it take her to cross the finish line from \(75 \mathrm{m}\) away? Is this reasonable?

Short Answer

Expert verified
It takes approximately -4.7095 seconds for the runner to cross the finish line, which is not a reasonable value since we cannot have negative time. This indicates an error in the problem statement or the runner cannot decelerate at the specified rate and reach the finish line simultaneously.

Step by step solution

01

Identify the given values

In the given problem, we have: - Initial distance from the finish line, s = 75 m; - Initial speed of the runner, u = 8 m/s; - Deceleration, a = -0.5 m/s² (since she is decelerating); - Final velocity, v = 0 m/s (as the runner has stopped at the finish line).
02

Use the equation of motion

Using the equation of motion \(v^2 = u^2 + 2as\), we can plug in the given values: \[0^2 = 8^2 + 2(-0.5)(75)\] Now we solve for the value of v^2.
03

Solve for v²

Solve the equation step-by-step: \[0 = 64 - 75\] \[75 = 64\] Now we find the difference of velocity squared: \[v^2 - u^2 = 75 - 64\] \[v^2 - u^2 = 11\]
04

Calculate time t using velocity and acceleration

Now, we can use the equation \(v = u + at\) to find the time taken (t): \[v - u = at\] Substituting the values, we get: \[\sqrt{v^2 - u^2} = -0.5 \times t\] \[\sqrt{11} = -0.5 \times t\] Now solve for t.
05

Solve for t

Divide by -0.5 on both sides to get the time t: \[t = \frac{\sqrt{11}}{-0.5}\] The obtained time is approximately equal to -4.7095. Since we get a negative value for the time, which is not possible in real-life scenarios, this indicates that there is an error in the problem statement or the runner cannot decelerate at the specified rate and reach the finish line simultaneously. The answer is not reasonable in this case.

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Key Concepts

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

Deceleration
Deceleration is what happens when an object slows down. In this exercise, the runner is slowing down while nearing the finish line. This is important to understand because it directly affects the time it will take her to reach the stop.
When we talk about deceleration, it is seen as a negative acceleration. This means the speed of the object decreases as it moves forward. In our exercise, the runner's deceleration is given as -0.5 m/s². This value tells us how much the speed reduces each second. So, the runner's speed decreases by 0.5 meters per second, every second.
  • Deceleration is a negative change in velocity per unit time.
  • It tells us how quickly an object is coming to a stop or reducing its speed.
Understanding deceleration helps us calculate how long it might take for any moving object, like our runner, to come to a full stop. This concept is used in various practical situations, such as planning safe stopping distances for vehicles.
Equations of motion
Equations of motion help us understand how objects move under the force of gravity and other influences, like deceleration. They allow us to predict future positions and velocities of moving objects using math.
In the given problem, one of these equations,
  • \[v^2 = u^2 + 2as\]
allows us to calculate aspects of the runner's movement as she decelerates. Here:
  • \(v\) is the final velocity (0 m/s, when she stops)
  • \(u\) is the initial velocity (8 m/s)
  • \(a\) is the acceleration, which is given as a negative value because it's deceleration (-0.5 m/s²)
  • \(s\) is the distance over which the runner decelerates (75 m)

This equation helps explain how these factors interact to bring the runner to a stop. The equations of motion are fundamental tools in kinematics, aiding us in understanding how different factors affect the movement of objects across different scenarios.
Velocity
Velocity is how fast something is moving and the direction it is moving in. In the context of this exercise, the runner's initial velocity is 8 m/s, which means she is traveling at a speed of 8 meters per second towards the finish line.
Velocity is different from speed. While speed gives us a number that tells us how fast something moves, velocity also adds the direction. This distinction is particularly significant when considering changes in motion, like our runner's deceleration.
  • Initial velocity is the speed at which the runner starts from a given point.
  • Final velocity occurs when the runner stops, reaching 0 m/s in this case.
  • Changes in velocity (due to deceleration in this case) affect how long it will take to cover a distance.
This concept of velocity is crucial for understanding motion scenarios. In problems like these, knowing the initial and final velocities can help calculate time, distance, and changes in the object's state of motion effectively.

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Most popular questions from this chapter

A well-thrown ball is caught in a well-padded mitt. If the acceleration of the ball is \(2.10 \times 10^{4} \mathrm{m} / \mathrm{s}^{2},\) and 1.85 \(\mathrm{ms}\left(1 \mathrm{ms}=10^{-3} \mathrm{s}\right)\) elapses from the time the ball first touches the mitt until it stops, what is the initial velocity of the ball?

(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of \(8.00 \mathrm{m} / \mathrm{s}\). (b) How long a time would it take to reach the ground if it is thrown straight down with the same speed?

The position of a particle moving along the \(x\) -axis varies with time according to \(x(t)=5.0 t^{2}-4.0 t^{3} \mathrm{m} .\) Find (a) the velocity and acceleration of the particle as functions of time, (b) the velocity and acceleration at \(t=2.0 \mathrm{s},(\mathrm{c})\) the time at which the position is a maximum, (d) the time at which the velocity is zero, and (e) the maximum position.

When analyzing the motion of a single object, what is the required number of known physical variables that are needed to solve for the unknown quantities using the kinematic equations?

There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. (a) How fast will it be going when it strikes the ground? (b) Assuming a reaction time of 0.300 s, how long a time will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is \(335.0 \mathrm{m} / \mathrm{s}\) on this day.

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