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Chapter 2: Problem 74

A runner of mass \(56.1 \mathrm{~kg}\) starts from rest and accelerates with a constant acceleration of \(1.23 \mathrm{~m} / \mathrm{s}^{2}\) until she reaches a velocity of \(5.10 \mathrm{~m} / \mathrm{s}\). She then continues running at this constant velocity. How long does the runner take to travel \(173 \mathrm{~m} ?\)

Short Answer

Expert verified
Answer: To find the total time taken by the runner to travel 173 meters, follow these steps: 1. Calculate the time for the acceleration phase using \(t = \frac{v - u}{a}\). 2. Calculate the distance traveled during acceleration using \(s = ut + \frac{1}{2} at^2\). 3. Calculate the remaining distance to be covered by subtracting the distance traveled during acceleration from 173 meters. 4. Calculate the time for the constant velocity phase using \(t = \frac{s}{v}\). 5. Add the time spent during the acceleration and constant velocity phases to find the total time required for the runner to travel 173 meters.

Step by step solution

01

Calculate the time for the acceleration phase

First, we need to find how long the runner takes to reach the constant velocity of \(5.10 \mathrm{~m} / \mathrm{s}\). We can use the equation \(v = u + at\), where \(v\) is the final velocity, \(u\) is the initial velocity (which is \(0\) since the runner starts from rest), \(a\) is the acceleration, and \(t\) is the time: \begin{equation} t = \frac{v - u}{a} \end{equation} Plugging in values: \(v = 5.10 \mathrm{~m/s}, u = 0 \mathrm{~m/s}\) and \(a = 1.23 \mathrm{~m/s^2}\), we'll get the time for the acceleration phase.
02

Calculate the distance traveled during acceleration

Next, we need to find the distance traveled during the acceleration phase. We'll use the equation \(s = ut + \frac{1}{2} at^2\), where \(s\) is the distance traveled, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time found in Step 1. Plugging in values: \(u = 0 \mathrm{~m/s}, a = 1.23 \mathrm{~m/s^2}\), and \(t\) from Step 1, we'll find the distance traveled during acceleration.
03

Calculate the distance remaining to be covered

Now, we need to find the remaining distance the runner needs to cover before reaching \(173 \mathrm{~m}\). We can do this by subtracting the distance traveled during acceleration (found in Step 2) from \(173 \mathrm{~m}\).
04

Calculate the time for the constant velocity phase

Now, we'll find the time required to cover the remaining distance at a constant velocity of \(5.10 \mathrm{~m/s}\). We can use the equation \(t = \frac{s}{v}\), where \(t\) is the time, \(s\) is the distance remaining (found in Step 3), and \(v\) is the constant velocity.
05

Calculate the total time the runner takes to travel \(173 \mathrm{~m}\)

Finally, we'll add the time spent during the acceleration phase (Step 1) and the time spent during the constant velocity phase (Step 4) to find the total time required for the runner to travel \(173 \mathrm{~m}\).

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Most popular questions from this chapter

A stone is thrown downward with an initial velocity of \(10.0 \mathrm{~m} / \mathrm{s}\). The acceleration of the stone is constant and has the value of the free-fall acceleration, \(9.81 \mathrm{~m} / \mathrm{s}^{2} .\) What is the velocity of the stone after \(0.500 \mathrm{~s} ?\)

A car traveling at \(25.0 \mathrm{~m} / \mathrm{s}\) applies the brakes and decelerates uniformly at a rate of \(1.2 \mathrm{~m} / \mathrm{s}^{2}\) a) How far does it travel in \(3.0 \mathrm{~s}\) ? b) What is its velocity at the end of this time interval? c) How long does it take for the car to come to a stop? d) What distance does the car travel before coming to a stop?

Two cars are traveling at the same speed, and the drivers hit the brakes at the same time. The deceleration of one car is double that of the other. By what factor does the time required for that car to come to a stop compare with that for the other car?

The Bellagio Hotel in Las Vegas, Nevada, is well known for its Musical Fountains, which use 192 HyperShooters to fire water hundreds of feet into the air to the rhythm of music. One of the HyperShooters fires water straight upward to a height of \(240 \mathrm{ft}\). a) What is the initial speed of the water? b) What is the speed of the water when it is at half this height on its way down? c) How long will it take for the water to fall back to its original height from half its maximum height?

An airplane starts from rest and accelerates at \(12.1 \mathrm{~m} / \mathrm{s}^{2}\). What is its speed at the end of a \(500 .-\mathrm{m}\) runway?
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