Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

An electron’s position is given by $\vec{r} = 3.00 t \hat{i} - 4.00 t^{2} \hat{j} + 2.00 \hat{k}$ , with $t$ in seconds and $\vec{r}$ in meters. (a)In unit-vector notation, what is the electron’s velocity $\hat{v} (t)$ ? At $t = 2.00 s$ , what is $\vec{v}$ (b) in unit-vector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the $x$ axis?

The velocity of electron is $(3.0 \hat{i} - 8.0 t \hat{j}) m / s$ .
The velocity of electron at t=2.0 s is $(3.0 \hat{i} - 16.0 t \hat{j}) m / s$ .
The magnitude of velocity of electron is 16.3 m/s.
The angle of velocity relative to the positive direction of x-axis is $- 79.4 °$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The position vector of electron, $\vec{r} = (3.0 t) \hat{i} - (4.0 t^{2}) \hat{j} + (2.0) \hat{k}$

Step 2: Understanding the velocity

The velocity is the rate of change of position of an object with respect to time. It is a vector quantity, which has magnitude as well as direction.

The expression for velocity in general can be written as:

$\vec{v} = \frac{d \vec{r}}{d t}$ ...(i)

Here, $\vec{r}$ is the position vector.

The magnitude of is given as:

$v = \sqrt{{v^{2}}_{x} + {v^{2}}_{y}}$ ....(ii)

Here, $v_{x}$ and $v_{y}$ are the x and y components of velocity.

The direction of $\vec{v}$ is given as:

$θ = \tan^{- 1} (\frac{v_{y}}{v_{x}})$ … (iii)

Step 3: (a) Determination of theelectron’s velocity

Using equation (i), the velocity of electron is calculated as:

$\vec{v} (t) = \frac{d [(3.0 t) \hat{i} - (4.0 t^{2}) \hat{j} + 2.0 \hat{k}]}{d t} = (3.0 \hat{i} - 8.0 t \hat{j}) m / s$

Thus, the velocity of electron is $(3.0 \hat{i} - 8.0 t \hat{j}) m / s$ .

Step 4: (b) Determination of theelectron’s velocity at

Substitute $t = 2.0 s$ in the above expression to find the velocity.

$\vec{v} = (3.0 \hat{i} - 8.0 \times 2.0 \hat{j}) m / s = (3.0 \hat{i} - 16.0 t \hat{j}) m / s$

Thus, the velocity of electron at t=2.0 s is $(3.0 \hat{i} - 16.0 t \hat{j}) m / s$ .

Step 5: (c) Determination of themagnitude of velocity

Using equation (ii), the magnitude of velocity is calculated as:

$v = \sqrt{{(3.0 m / s)}^{2} + {(16.0 m / s)}^{2}} = 16.3 m / s$

Thus, the magnitude of velocity of electron is 16.3 m/s .

Step 6: (d) Determination ofthedirection of velocity

Using equation (iii), the direction of electron’s velocity is calculated as:

$θ = \tan^{- 1} (\frac{- 16.0 m / s}{3.0 m / s}) = \tan^{-} (- 5.333) = - 79.4 °$

Thus, the angle of velocity relative to the positive direction of x-axis is $- 79.4 °$ .

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Understanding the velocity

Step 3: (a) Determination of theelectron’s velocity

Step 4: (b) Determination of theelectron’s velocity at

Step 5: (c) Determination of themagnitude of velocity

Step 6: (d) Determination ofthedirection of velocity

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Fundamentals Of Physics

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Step 4: (b) Determination of theelectron’s velocity at

Step 5: (c) Determination of themagnitude of velocity

Step 6: (d) Determination ofthedirection of velocity

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