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Chapter 2: Problem 55

A typical male sprinter can maintain his maximum acceleration for \(2.0 \mathrm{~s}\) and his maximum speed is \(10 \mathrm{~m} / \mathrm{s}\). After reaching this maximum speed, his acceleration becomes zero and then he runs at constant speed. Assume that his acceleration is constant during the first \(2.0 \mathrm{~s}\) of the race, that he starts from rest, and that he runs in a straight line. (a) How far has the sprinter run when he reaches his maximum speed? (b) What is the magnitude of his average velocity for a race of the following lengths: (i) \(50.0 \mathrm{~m}\) (ii) \(100.0 \mathrm{~m},\) (iii) \(200.0 \mathrm{~m} ?\)

Short Answer

Expert verified
(a) 10 meters; (b) (i) ~8.33 m/s, (ii) ~9.09 m/s, (iii) ~9.52 m/s.

Step by step solution

01

Understanding Initial Conditions

The sprinter starts from rest, which means his initial velocity \( v_0 = 0 \) m/s and begins to accelerate at a constant rate up to his maximum speed of \( 10 \) m/s in \( 2 \) seconds.
02

Calculate Constant Acceleration

Using the formula \( v = u + at \), where \( v = 10 \) m/s, \( u = 0 \) m/s (initial velocity), and \( t = 2 \) s, we solve for \( a \) (acceleration): \( 10 = 0 + a \times 2 \Rightarrow a = \frac{10}{2} = 5 \) m/s虏.
03

Determine Distance to Maximum Speed

To find out how far the sprinter runs during the acceleration phase, use the equation \( s = ut + \frac{1}{2}at^2 \). With \( u = 0 \), \( a = 5 \) m/s虏, and \( t = 2 \) s, we calculate:\[ s = 0 \times 2 + \frac{1}{2} \times 5 \times (2)^2 = 10 \text{ meters} \]
04

Average Velocity for 50 Meter Race

For a race shorter than \( 2 \) seconds, sprinter reaches max speed partway. Since he reaches 10 m/s in 10 meters, mix velocities: - Accelerates(first 10m) for \( t=2 \) s.- Constant speed (40m) for \( t=\text{(2+(40/10))} \). Average velocity: \[ \frac{50}{\text{total time}} = \frac{50}{6} \approx 8.33 \text{ m/s} \]
05

Average Velocity for 100 Meter Race

For a 100 m race, the sprinter runs 10 m accelerating in \( 2 \) s, and 90 m at maximum speed (\( 10 \) m/s) in \( 9 \) seconds. Total time is \( 2+9=11 \) seconds, thus: \[ v_{avg} = \frac{100}{11} \approx 9.09 \text{ m/s} \]
06

Average Velocity for 200 Meter Race

For 200 m, sprinter accelerates 10 m in 2s and covers 190 m at max speed (10 m/s), so: - Time for acceleration is 2s.- Constant velocity: \( \frac{190}{10} = 19 \)s.Total time: \( 2+19=21 \) seconds. Average velocity: \[ v_{avg} = \frac{200}{21} \approx 9.52 \text{ m/s} \]

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Key Concepts

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

Acceleration
When discussing motion in kinematics, acceleration is a key concept. Acceleration describes how quickly an object changes its velocity. In this exercise, the sprinter demonstrates a constant acceleration phase at the start of his run. Starting from rest, the sprinter accelerates to reach a velocity of 10 m/s in 2 seconds.
To find the acceleration, we use the equation:
  • \( v = u + at \)
where:
  • \( v \) is the final velocity (10 m/s),
  • \( u \) is the initial velocity (0 m/s),
  • \( a \) is the acceleration,
  • \( t \) is the time (2 seconds).
By rearranging and solving for \( a \), we find that \( a = 5 \) m/s虏. This value indicates a constant acceleration, meaning each second the sprinter increases his speed by 5 m/s until reaching his maximum speed.
Constant Velocity
Once the sprinter reaches his maximum speed of 10 m/s, he transitions from accelerating to moving at a constant velocity. Constant velocity means that the speed is unchanging, and no acceleration is involved. In this state, the sprinter covers equal distances in equal time intervals.
After the initial 10 meters covered with acceleration, the sprinter maintains the same speed for the remainder of the race. The speed does not increase or decrease as time passes. For this exercise, his race performance involves different segments:
  • First 10 meters: accelerating
  • Remaining distance (e.g., 40 meters in a 50-m race): constant velocity
This method makes it easier to calculate distances and times during phases over longer distances.
Average Velocity
Average velocity is an important concept in understanding motion over a distance. It is defined as the total distance traveled divided by the total time taken. In kinematics, computing average velocity gives insight into the overall speed of a moving object along its path.
In this exercise, the sprinter's average velocity changes depending on the race length:
  • For a 50-meter race: The average velocity is approximately 8.33 m/s, calculated using both acceleration and constant velocity phases.
  • For a 100-meter race: The average velocity increases to about 9.09 m/s because the sprinter spends longer time at max speed.
  • For a 200-meter race: The average velocity is further optimized to around 9.52 m/s because the constant velocity phase dominates the time span.
Calculating the average velocity in multi-phase runs requires considering the time spent in both the acceleration and constant velocity phases.

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Most popular questions from this chapter

Catching the Bus. A student is running at her top speed of 5.0 \(\mathrm{m} / \mathrm{s}\) to catch a bus, which is stopped at the bus stop. When the student is still 40.0 \(\mathrm{m}\) from the bus, it starts to pull away, moving with a constant acceleration of 0.170 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) For how much time and what distance does the student have to run at 5.0 \(\mathrm{m} / \mathrm{s}\) before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling? (c) Sketch an \(x\) -t graph for both the student and the bus. Take \(x=0\) at the initial position of the student. (d) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus traveling at this point? (e) If the student's top speed is 3.5 \(\mathrm{m} / \mathrm{s},\) will she catch the bus? (f) What is the minimum speed the student must have to just catch up with the bus? For what time and what distance does she have to run in that case?

A rocket carrying a satellite is accelerating straight up from the earth's surface. At 1.15 s after liftoff, the rocket clears the top of its launch platform, 63 \(\mathrm{m}\) above the ground. After an additional \(4.75 \mathrm{s},\) it is 1.00 \(\mathrm{km}\) above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75 -s part of its flight and (b) the first 5.90 s of its flight.

If a pilot accelerates at more than \(4 g,\) he begins to "gray out" but doesn't completely lose consciousness. (a) Assuming constant acceleration, what is the shortest time that a jet pilot starting from rest can take to reach Mach 4 (four times the speed of sound) without graying out? (b) How far would the plane travel during this period of acceleration? (Use 331 \(\mathrm{m} / \mathrm{s}\) for the speed of sound in cold air.)

A small rock is thrown vertically upward with a speed of 18.0 \(\mathrm{m} / \mathrm{s}\) from the edge of the roof of a 30.0 -m-tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected. (a) What is the speed of the rock just before it hits the street? (b) How much time elapses from when the rock is thrown until it hits the street?

A bird is flying due east. Its distance from a tall building is given by \(x(t)=28.0 \mathrm{m}+(12.4 \mathrm{m} / \mathrm{s}) t-\left(0.0450 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}\) What is the instantaneous velocity of the bird when \(t=8.00 \mathrm{s}?\)
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