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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 train traveling at \(40.0 \mathrm{~m} / \mathrm{s}\) is headed straight toward another train, which is at rest on the same track. The moving train decelerates at \(6.0 \mathrm{~m} / \mathrm{s}^{2},\) and the stationary train is \(100.0 \mathrm{~m}\) away. How far from the stationary train will the moving train be when it comes to a stop?

A girl is riding her bicycle. When she gets to a corner, she stops to get a drink from her water bottle. At that time, a friend passes by her, traveling at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s}\). a) After \(20 \mathrm{~s}\), the girl gets back on her bike and travels with a constant acceleration of \(2.2 \mathrm{~m} / \mathrm{s}^{2} .\) How long does it take for her to catch up with her friend? b) If the girl had been on her bike and rolling along at a speed of \(1.2 \mathrm{~m} / \mathrm{s}\) when her friend passed, what constant acceleration would she need to catch up with her friend in the same amount of time?

An object is thrown upward with a speed of \(28.0 \mathrm{~m} / \mathrm{s}\). How long does it take it to reach its maximum height?

The position of an object as a function of time is given as \(x=A t^{3}+B t^{2}+C t+D .\) The constants are \(A=2.1 \mathrm{~m} / \mathrm{s}^{3}\) \(B=1.0 \mathrm{~m} / \mathrm{s}^{2}, C=-4.1 \mathrm{~m} / \mathrm{s},\) and \(D=3 \mathrm{~m}\) a) What is the velocity of the object at \(t=10.0 \mathrm{~s}\) ? b) At what time(s) is the object at rest? c) What is the acceleration of the object at \(t=0.50 \mathrm{~s} ?\) d) Plot the acceleration as a function of time for the time interval from \(t=-10.0 \mathrm{~s}\) to \(t=10.0 \mathrm{~s}\).

The minimum distance necessary for a car to brake to a stop from a speed of \(100.0 \mathrm{~km} / \mathrm{h}\) is \(40.00 \mathrm{~m}\) on a dry pavement. What is the minimum distance necessary for this car to brake to a stop from a speed of \(130.0 \mathrm{~km} / \mathrm{h}\) on dry pavement?
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