Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

Answers without the blur.

Just sign up for free and you're in.

Short Answer

A graph of v(t) is shown for a world-class track sprinter in a \({\bf{100}} - {\bf{m}}\) race. (See Figure 2.67). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at \(t = {\rm{ }}{\bf{5}}{\rm{ }}{\bf{s}}\)? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

(a) \(6 m/s\)

(b) \(12{\rm{ m/s}}\)

(d) \(10.3 s\)

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Graph and its analysis

To analyses movement, motion graphs can be employed.

The mathematical approaches and graphical solutions for determining motion equations are the same.

The slope of a graph of displacement x vs. time t represents velocity v.

The inclination of a graph of velocity v vs. time t is called acceleration.

Average velocity, instantaneous velocity, and acceleration can all be calculated using graphs.

Average velocity for the first 4 s

a) Here if we carefully look into the figure, in your text book lets take any two points on the line

\(\begin{array}{c}average\,velocity = \frac{{{v_1} + {v_2}}}{2}\\ = \frac{{0 + 12}}{2}\\ = 6\,m/s\end{array}\)

Hence the average velocity of the body up to 4 s is 6 m/s

instantaneous velocity at t = 5 s

Hence when the time is 5s the velocity is constantly 12 m/s.

Average acceleration between 0 to 4 s

(c)

Here if we carefully look into the figure, in your text book let’s take any two points on the line

Coordinate \(\left( {0,0} \right)\) and \(\left( {4,12} \right)\)

Slop of the line will be

\(\begin{array}{l}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_2}}}\\Here\,the\,coordinates\,are\,(0,0)\,\,\,and\,\,\,(4,12)\\m = \frac{{12 - 0}}{{4 - 0}}\\m = 3\,m/{s^2}\end{array}\)

Hence the acceleration is approximately to the value\(3 m/{s^2}\).

Time for the race

(d) For the first \(4{\rm{ s}}\)

Distance travelled is equal to area under the curve

Distance = \(\frac{1}{2} \times 4 \times 12 = 24m\)

He can run the remaining distance by

\(\begin{array}{c}t = \frac{{distance}}{{velocity}} = \frac{{76}}{{12}}\\ = 6.3 s\end{array}\)

Total time of the sprint is\({t_{total}} = 4 + 6.3 = 10.3 s\)

Find Study Materials for

Create Study Materials

Select your language

College Physics (Urone)

Answers without the blur.

Short Answer

Step by Step Solution

Graph and its analysis

Average velocity for the first 4 s

instantaneous velocity at t = 5 s

Average acceleration between 0 to 4 s

Time for the race

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

College Physics (Urone)

Answers without the blur.

Short Answer

Step by Step Solution

Graph and its analysis

Average velocity for the first 4 s

instantaneous velocity at t = 5 s

Average acceleration between 0 to 4 s

Time for the race

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics