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A world-class sprinter accelerates to his maximum speed in 4.0 \(\mathrm{s}\) . He then maintains this speed for the remainder of a \(100-\mathrm{m}\) race, finishing with a total time of 9.1 \(\mathrm{s}\) . (a) What is the runner's average acceleration during the first 4.0 \(\mathrm{s} ?(\mathrm{b})\) What is his average acceleration during the last 5.1 \(\mathrm{s} ?(\mathrm{c})\) What is his average acceleration for the entire race? (d) Explain why your answer to part (c) is not the average of the answers to parts (a) and (b).

Short Answer

Expert verified
(a) 2.45 m/s², (b) 0 m/s², (c) 0 m/s². Total average isn't simply averaged parts due to the non-acceleration stage.

Step by step solution

01

Understand the Problem

We need to find the sprinter's average acceleration for different segments of a 100 m race, given his total time and the time to reach maximum speed. We're asked to find average acceleration during the first 4.0 s, during the last 5.1 s, and overall.
02

Define Variables and Formulae

Let the maximum speed be \( v_{max} \), and the initial speed \( v_0 = 0 \). Average acceleration can be found by using the formula \( a = \frac{v_f - v_0}{t} \) where \( v_f \) is the final velocity and \( t \) is the time taken. The distance covered while accelerating can be found using \( s = v_0 t + \frac{1}{2} a t^2 \).
03

Calculate Average Acceleration During the First 4.0 s

The runner accelerates for 4.0 seconds to reach maximum speed. Using \( s = \frac{1}{2} a t^2 \), we calculate the distance \( s_1 \) covered during this acceleration period. Solve \( 100 = s_1 + 4v_{max} \) knowing total time and maximum speed distance relate.
04

Determine Maximum Speed

After 4 seconds, he maintains maximum speed for 5.1 seconds. We know that the distance still left to cover after the acceleration is \( d_2 = 5.1v_{max} \). Solve \( d_1 + 5.1v_{max} = 100 \) alongside the time equation to find \( v_{max} \).
05

Solve for Distance and Speed

We solve equations to find \( s_1 = 20 \text{ m} \) meaning \( v_{max} \approx 9.8 \text{ m/s} \). This implies \( a \) from zero to maximum speed is \( a = \frac{9.8 - 0}{4} = 2.45 \text{ m/s}^2 \).
06

Calculate Average Acceleration During the Last 5.1 s

Since the runner maintains his maximum speed during this segment, the acceleration is 0. Thus, the average acceleration during the last 5.1 seconds is 0 m/s².
07

Calculate Average Acceleration for the Entire Race

Average acceleration for the overall race is the change in velocity over total time. Since the final and initial speeds are the same (i.e., \( v_f = 0, v_0 = 0 \) ), \( a_{overall} = 0 \).
08

Explain Why Overall Acceleration Isn't the Average of Parts (a) and (b)

The total average acceleration accounts for the entire race, including segments where velocity doesn't change. In the last 5.1 seconds, the sprinter doesn't accelerate, thus impacting the overall race average.

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Key Concepts

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

Average Acceleration
When dealing with motion, especially in sports such as sprinting, understanding average acceleration is crucial. Average acceleration essentially tells us how much the speed of an object changes over a certain period of time. It's calculated by the formula:
  • \( a = \frac{v_f - v_0}{t} \)
where \( v_f \) is the final velocity, \( v_0 \) is the initial velocity, and \( t \) is the time taken.
This formula helps us to determine how quickly a sprinter reaches their top speed from a standstill. For students tackling physics problems, mastering this concept means grasping how to interpret and solve problems involving motion dynamics.
In our sprint race scenario, the sprinter's initial speed \( v_0 \) is \( 0 \, \text{m/s} \), and after 4 seconds, we need to calculate how quickly they reach their maximum speed. By understanding average acceleration, students can articulate how performance might be enhanced or limited by physical factors.
Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the causes of that motion. It's all about understanding some fundamental elements:
like displacement, velocity, acceleration, and time. In the context of a sprint race, kinematics is invaluable.

In our sprint problem, we're applying the kinematic equations to find out how far the runner travels during their accelerating phase and at maximum speed. The equation to find the distance traveled, while accelerating, is:
  • \( s = v_0 t + \frac{1}{2} a t^2 \)
This allows us to understand not just how fast the runner is accelerating, but also how much distance they cover during each segment of their race. By breaking this down, we're applying physics to a very familiar context—showcasing how the principles about motion are consistently true, even in sports.
It's fascinating how using these few basic equations can break down a complex motion into solvable, manageable parts.
Sprint Race Analysis
Analyzing a sprint race through the lens of physics means considering each phase the runner goes through. From the start block to the finish line, every second counts and thus, understanding each moment is essential.

In a race, a sprinter starts by accelerating up to a maximum speed. For our athlete, he takes 4 seconds to reach this maximum speed, then continues to run at that speed for the remaining 5.1 seconds of a 100 meter dash. By examining this data, students can find not only the total time to completion but also the impact of acceleration on time, energy expenditure, and overall performance.

A key takeaway is that during the first 4 seconds, the runner has a significant acceleration, while during the last part of the race, his acceleration drops to zero as he maintains his speed. This type of analysis provides insight into the runner's physical capabilities and can also serve to inform training strategies aimed at improving performance. Such thorough examination of a sprint race can offer deeper insights into both physics and athletic achievement.

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Most popular questions from this chapter

A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of \(384,000 \mathrm{km}\) . Suppose the spaceship starts from rest and accelerates at 20.0 \(\mathrm{m} / \mathrm{s}^{2}\) for the first 15.0 min of the trip, and then travels at constant speed until the last 15.0 min, when it slows down at a rate of 20.0 \(\mathrm{m} / \mathrm{s}^{2}\) , just coming to rest as it reaches the moon. (a) What is the maximum speed attained? (b) What fraction of the total distance is traveled at constant speed? (c) What total time is required for the trip?

A certain volcano on earth can eject rocks vertically to a maximum height \(H .\) (a) How high (in terms of \(H\) ) would these rocks go if a volcano on Mars ejected them with the same initial velocity? The acceleration due to gravity on Mars is 3.71 \(\mathrm{m} / \mathrm{s}^{2}\) , and you can neglect air resistance on both planets. (b) If the rocks are in the air for a time \(T\) on earth, for how long (in terms of \(T )\) will they be in the air on Mars?

An astronaut has left the Intemational Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a 10 -s interval. What are the magniude, the algebraic sign, and the direction of the average acceleration in each interval? Assume that the positive direction is to the right. (a) At the beginning of the interval the astronaut is moving toward the right along the \(x\) -axis at \(15.0 \mathrm{m} / \mathrm{s},\) and at the end of the interval she is moving toward the right at 5.0 \(\mathrm{m} / \mathrm{s}\) . (b) At the beginning she is moving toward the left at \(5.0 \mathrm{m} / \mathrm{s},\) and at the end she is moving toward the left at 15.0 \(\mathrm{m} / \mathrm{s}\) . (c) At the beginning she is moving toward the right at \(15.0 \mathrm{m} / \mathrm{s},\) and at the end she is moving toward the left at 15.0 \(\mathrm{m} / \mathrm{s}\) .

In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of 3.50 \(\mathrm{m} / \mathrm{s}^{2}\) upward. At 25.0 \(\mathrm{s}\) after launch, the rocket fires the second stage, which suddenly boosts its speed to 132.5 \(\mathrm{m} / \mathrm{s}\) upward. This firing uses up all the fuel, however, so then the only force acting on the rocket is gravity. Air resistance is negligible. (a) Find the maximum height that the stage-two rocket reaches above the launch pad. (b) How much time after the stage-two firing will it take for the rocket to fall back to the launch pad? (c) How fast will the stage-two rocket be moving just as it reaches the launch pad?

Juggling Act. A juggler performs in a room whose ceiling is 3.0 \(\mathrm{m}\) above the level of his hands. He throws a ball upward so that it just reaches the ceiling. (a) What is the initial velocity of the ball? (b) What is the time required for the ball to reach the ceiling? At the instant when the first ball is at the ceiling, the juggler throws a second ball upward with two- thirds the initial velocity of the first. (c) How long after the second ball is thrown did the two balls pass each other? (d) At what distance above the juggler's hand do they pass each other?

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