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Chapter 2: Problem 53

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 he reaches 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 these lengths: (i) \(50.0 \mathrm{~m} ;\) (ii) \(100.0 \mathrm{~m}\); (iii) \(200.0 \mathrm{~m} ?\)

Short Answer

Expert verified
(a) The sprinter has run 10 meters when he reaches his maximum speed. (b) The magnitude of his average velocity for a race of these lengths are (i) 6.25 m/s, (ii) 7.69 m/s, (iii) 8.33 m/s.

Step by step solution

01

Calculate the distance traveled during acceleration

Using the formula \(s = ut + \frac{1}{2}a(t^2)\), where \(u\) is the initial velocity (0 m/s because the sprinter starts from rest), \(t\) is the time of 2 seconds, and \(a\) is the acceleration (which can be found by \(v = u + at\), where \(v\) is 10 m/s and \(u\) is 0 m/s), gives \(a = 5m/s^2\). Substituting these values into the equation, we get \(s = 0 * 2 + \frac{1}{2} * 5 * (2^2) = 10 m\).
02

Calculate average velocity over different distances

The sprinter accelerates for the first 2 seconds and then runs at constant speed. Therefore the total time to cover a distance is \(2 + \frac{{(d -10)}}{{10}}\) where \(d\) is the total distance. The average velocity is then \(\frac{d}{{2 + \frac{{(d -10)}}{{10}}}}\)
03

Substitute the distances into the formula

(i) For 50.0 m, the average velocity is \(\frac{50}{{2 + \frac{{(50 -10)}}{{10}}}} = 6.25 m/s\). (ii) For 100.0 m, the average velocity is \(\frac{100}{{2 + \frac{{(100 -10)}}{{10}}}} = 7.69 m/s\). (iii) For 200.0 m, the average velocity is \(\frac{200}{{2 + \frac{{(200 -10)}}{{10}}}} = 8.33 m/s\).

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Key Concepts

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

Constant Acceleration
In the realm of kinematics, constant acceleration is a condition where an object's rate of change of velocity remains steady over time. It's crucial in understanding motion scenarios like the one faced by our sprinter, who accelerates steadily for a set duration before he hits his max speed.

Constant acceleration permits us to use simplified equations because the acceleration, by definition, doesn't vary. This predictability is beneficial when solving problems related to motion because it allows us to use a formula to find distance covered, such as \( s = ut + \frac{1}{2}at^2 \), where \( s \) represents the distance traveled, \( u \) is the initial velocity, \( t \) is time, and \( a \) is our constant acceleration.
Average Velocity
Average velocity is the total displacement divided by the total time taken for the displacement. It's an essential concept for understanding how fast an object is moving on average, rather than at any specific instant.

When our sprinter reaches his top speed, he no longer accelerates and his instant velocity is constant. However, since he did not move at this speed for the entire race, we must calculate his average velocity to get a full picture of his performance. This value represents the constant velocity at which the sprinter would need to run to cover the given distance in the same time frame.
Kinematic Equations
Kinematic equations allow us to predict the future position of an object in motion, based on its current state, assuming that the motion takes place in a straight line and is accurately described by the parameters in the equations.

In the context of constant acceleration, a set of four key kinematic equations can be used to calculate various attributes of motion, such as velocity, acceleration, time, and displacement. These formulas are powerful tools because they interlink all these attributes and enable us to solve for one when we have information about the others. For our exercise, we utilized the equation \( s = ut + \frac{1}{2}at^2 \) to determine the sprinter's displacement during his acceleration phase.

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Most popular questions from this chapter

A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline, and its speed after it has traveled \(6.80 \mathrm{~m}\) to the bottom of the incline is \(3.80 \mathrm{~m} / \mathrm{s}\). What is the speed of the block when it is \(3.40 \mathrm{~m}\) from the top of the incline?

The human body can survive an acceleration trauma incident (sudden stop) if the magnitude of the acceleration is less than \(250 \mathrm{~m} / \mathrm{s}^{2}\). If you are in an automobile accident with an initial speed of \(105 \mathrm{~km} / \mathrm{h}(65 \mathrm{mi} / \mathrm{h})\) and are stopped by an airbag that inflates from the dashboard, over what minimum distance must the airbag stop you for you to survive the crash?

You are climbing in the High Sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of this cliff, you drop a rock from the top; 8.00 s later you hear the sound of the rock hitting the ground at the foot of the cliff. (a) If you ignore air resistance, how high is the cliff if the speed of sound is \(330 \mathrm{~m} / \mathrm{s} ?\) (b) Suppose you had ignored the time it takes the sound to reach you. In that case, would you have overestimated or underestimated the height of the cliff? Explain.

A sprinter runs a \(100 \mathrm{~m}\) dash in \(12.0 \mathrm{~s}\). She starts from rest with a constant acceleration \(a_{x}\) for \(3.0 \mathrm{~s}\) and then runs with constant speed for the remainder of the race. What is the value of \(a_{x} ?\)

The acceleration of a bus is given by \(a_{x}(t)=\alpha t,\) where \(\alpha=1.2 \mathrm{~m} / \mathrm{s}^{3} .\) (a) If the bus's velocity at time \(t=1.0 \mathrm{~s}\) is \(5.0 \mathrm{~m} / \mathrm{s}\) what is its velocity at time \(t=2.0 \mathrm{~s} ?\) (b) If the bus's position at time \(t=1.0 \mathrm{~s}\) is \(6.0 \mathrm{~m},\) what is its position at time \(t=2.0 \mathrm{~s} ?(\mathrm{c})\) Sketch \(a_{y}-t\) \(v_{y}-t,\) and \(x-t\) graphs for the motion.
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