Chapter 1, Problem 15P

To determine

The average speed over the different time intervals.

Explanation of Solution

Given:

The three intervals of the motion is $1.0\times {10}^{-1}\text{s}$ , $1.0\times {10}^{-2}\text{s}$ and $1.0\times {10}^{-3}\text{s}$ .

The value of constant $c$ is $1.0\text{m}/{\text{s}}^3$ and initial time is $1.0000\text{ s}$ . $$

Formula Used:

Speed is the distance covered in unit time by an object. Instantaneous speed is the speed of object at any instant.

Write the expression for final time.

  $t^{\prime }=t+\Delta t$ ...... (I)

Here, $t$ is the initial time, $\Delta t$ is the time interval and $t^{\prime }$ is the final interval.

Write the expression for speed of an object for time interval.

  $v=\frac{s^{\prime }-s}{t^{\prime }-t}$

Here, $s$ is the final distance, $s^{\prime }$ is the final speed and $v$ is the speed of object.

Substitute $c{t^{\prime }}^3$ for $s^{\prime }$ and $c{t}^3$ for $s$ in above expression.

  $v=\frac{c\left({t^{\prime }}^3-t^3\right)}{t^{\prime }-t}$ ...... (II)

Here, $c$ is the constant given.

Write the expression for instantaneous speed.

  $v_i=\frac{ds}{dt}$

Here, $v_i$ is the instantaneous speed, $ds$ is the small change in displacement and $dt$ is the small change in time.

Substitute $c{t}^3$ for $s$ in above expressionand simplify.

  $v_i=3c{t}^2$ ...... (III)

Calculation:

For first time interval:

Substitute $1.0\times {10}^{-1}\text{s}$ for $\Delta t_1$ and $1.0000\text{ s}$ for $t$ in equation (I).

  $\begin{array}{c}{t^{\prime }}_1=1.0000\text{ s}+0.1000\text{s}\\ =1.1000\text{s}\end{array}$

Substitute $1.1000\text{s}$ for ${t^{\prime }}_1$ and $1.0000\text{ s}$ for $t$ in equation (II).

  $\begin{array}{c}{v}_1=\frac{\left(1.0\text{m}/{\text{s}}^3\right)\left({\left(1.1000\text{s}\right)}^3-{\left(1.0000\text{s}\right)}^3\right)}{\left(1.1000\text{s}\right)-\left(1.0000\text{s}\right)}\\ =3.3100\text{m}/\text{s}\end{array}$

For second time interval:

Substitute $1.0\times {10}^{-2}\text{s}$ for $\Delta t_2$ and $1.0000\text{ s}$ for $t$ in equation (I).

  $\begin{array}{c}{t^{\prime }}_2=1.0000\text{ s}+0.0100\text{s}\\ =1.0100\text{s}\end{array}$

Substitute $1.0100\text{s}$ for ${t^{\prime }}_2$ and $1.0000\text{ s}$ for $t$ in equation (II).

  $\begin{array}{c}{v}_2=\frac{\left(1.0\text{m}/{\text{s}}^3\right)\left({\left(1.0100\text{s}\right)}^3-{\left(1.0000\text{s}\right)}^3\right)}{\left(1.0100\text{s}\right)-\left(1.0000\text{s}\right)}\\ =3.0301\text{m}/\text{s}\end{array}$

For third time interval:

Substitute $1.0\times {10}^{-3}\text{s}$ for $\Delta t_3$ and $1.0000\text{ s}$ for $t$ in equation (I).

  $\begin{array}{c}{t^{\prime }}_2=1.0000\text{ s}+0.0010\text{s}\\ =1.0010\text{s}\end{array}$

Substitute $1.0010\text{s}$ for ${t^{\prime }}_3$ and $1.0000\text{ s}$ for $t$ in equation (II).

  $\begin{array}{c}{v}_3=\frac{\left(1.0\text{m}/{\text{s}}^3\right)\left({\left(1.0010\text{s}\right)}^3-{\left(1.0000\text{s}\right)}^3\right)}{\left(1.0010\text{s}\right)-\left(1.0000\text{s}\right)}\\ =3.0030\text{m}/\text{s}\end{array}$

Substitute $1.0\text{m}/{\text{s}}^3$ for $c$ and $1.0000\text{ s}$ for $t$ in equation (III).

  $\begin{array}{c}{v}_i=3\left(1.0\text{m}/{\text{s}}^3\right){\left(1.0000\text{ s}\right)}^2\\ =3\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the average speed in given three intervals is $3.3100\text{m}/\text{s}$ , $3.0301\text{m}/\text{s}$ and $3.0030\text{m}/\text{s}$ ; the instantaneous speed is $3\text{m}/\text{s}$ .