Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

A $2.0 kg$ particle moves along an x axis, being propelled by a variable force directed along that axis. Its position is given by role="math" localid="1657014461331" $x = 3.0 m + (4.0 m / s) t + c t^{2} - (2.0 m / s) t^{3}$ , with x in meters and t in seconds. The factor c is a constant. At the force on the particle has a magnitude of $36 N$ and is in the negative direction of the axis. What is c?

The value of constant c is $+ 9.0$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

Mass of particle, $m = 2.0 kg$
Position of a particle in expression, $x = 3.0 + 4 t + c t^{2} - 2 t^{3}$
Magnitude of force at t = 3.0 s, $F = - 36 N$ which is in a negative direction

Step 2: Understanding the concept of motion

The acceleration of the object is equal to the time rate of change of velocity. The acceleration can be found by differentiating the velocity vector with respect to time or differentiating the displacement vector twice with respect to time.

As we are given the function for the position of a particle if we differentiate that for time, we will get twice the acceleration. As we have got the value of force at a particular time, using Newton’s 2nd law, we can find the value for c.

Formula:

The force on a body due to Newton’s second law, $\vec{F} = m \vec{a}$ (i)

Here, $\vec{F}$ is net force, is mass of the object, and $\vec{a}$ is the acceleration.

The acceleration of a body in motion, $a = \frac{d^{2} x}{d t^{2}}$ (ii)

Step 3: Calculation of the value of constant c

Taking the derivative of the given position equation $x = 3.0 + 4 t + c t^{2} - 2 t^{3}$ with respect to time, we get the velocity as follows:

role="math" localid="1657015128101" $v = 0 + 4 + 2 c t - 6 t^{2} . . . . . . . . . . . . . . (a)$

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Fundamentals Of Physics

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Step 3: Calculation of the value of constant c

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