Chapter 2, Problem 111P

(a)

To determine

The velocity as the function of time for the given time interval.

Answer to Problem 111P

The velocity as the function of time interval is $v\left(t\right)=\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^2+9.5\text{m}/\text{s}$ .

Explanation of Solution

Given:

The acceleration of the particle is given by $a_x=\left(0.20\text{m}/{\text{s}}^{\text{3}}\right)t$ .

The velocity of particle at $t=0$ is $9.5\text{m}/\text{s}$ .

The position of particle at $t=0$ is $-5.0\text{m}$ .

Formula used:

Write expression for the acceleration of the particle.

  $\frac{dv\left(t\right)}{dt}=a\left(t\right)$

Here, $a$ is the acceleration of the particle and $\frac{dv\left(t\right)}{dt}$ is the rate of change of velocity.

  $dv\left(t\right)=a\left(t\right)\cdot dt$

Substitute $\left(0.20\text{m}/{\text{s}}^{\text{3}}\right)$ for $a\left(t\right)$ in above expression and integrate.

  ${\displaystyle \int dv}\left(t\right)={\displaystyle \int \left(0.20\text{m}/{\text{s}}^{\text{3}}\right)t\cdot dt}$

Simplify above expression.

  $v\left(t\right)=\frac{\left(0.20\text{m}/{\text{s}}^{\text{3}}\right)}{2}{t}^2+C$............. (1)

Calculation:

Substitute $0$ for $t$ in equation (1).

  $\begin{array}{c}v\left(0\right)=\frac{\left(0.20\text{m}/{\text{s}}^{\text{3}}\right)}{2}{\left(0\right)}^2+C\\ C=v\left(0\right)\end{array}$

Substitute $9.5\text{m}/\text{s}$ for $v\left(0\right)$ in above expression.

  $C=9.5\text{m}/\text{s}$

Substitute $9.5\text{m}/\text{s}$ for $C$ in equation (1).

  $\begin{array}{c}v\left(t\right)=\frac{\left(0.20\text{m}/{\text{s}}^{\text{3}}\right)}{2}{t}^2+9.5\text{m}/\text{s}\\ =\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^2+9.5\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the velocity as the function of time interval is $v\left(t\right)=\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^2+9.5\text{m}/\text{s}$ .

(b)

To determine

The position as function of time for the given time interval.

Answer to Problem 111P

The position as function of time for the given interval is $x\left(t\right)=\frac{\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^3}{3}+\left(9.5\text{m}/\text{s}\right)t-5.0\text{m}$ .

Explanation of Solution

Given:

The acceleration of the particle is given by $a_x=\left(0.20\text{m}/{\text{s}}^{\text{3}}\right)t$ .

The velocity of particle at $t=0$ is $9.5\text{m}/\text{s}$ .

The position of particle at $t=0$ is $-5.0\text{m}$ .

Formula used:

Write expression for the velocity of the particle as function of time for given time period.

  $v\left(t\right)=\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^2+9.5\text{m}/\text{s}$

Write expression for velocity of the particle.

  $\frac{dx\left(t\right)}{dt}=v\left(t\right)$

Here, $v$ is the velocity of the particle and $\frac{dx}{dt}$ is the rate of change of position of the particle.

Rearrange above expression for $dx$ .

  $dx\left(t\right)=v\left(t\right)dt$  ........(2)

Calculation:

Substitute $\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^2+9.5\text{m}/\text{s}$ for $v\left(t\right)$ in equation (2) and integrate.

  ${\displaystyle \int dx\left(t\right)}={\displaystyle \int \left[\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^2+9.5\text{m}/\text{s}\right]}dt$

Simplify above expression.

  $x\left(t\right)=\frac{\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^3}{3}+\left(9.5\text{m}/\text{s}\right)t+D$

  ........(3)

Substitute $0$ for $t$ in above expression.

  $\begin{array}{l}x\left(0\right)=\frac{\left(0.10\text{m}/{\text{s}}^{\text{3}}\right){\left(0\right)}^3}{3}+\left(9.5\text{m}/\text{s}\right)\left(0\right)+D\\ x\left(0\right)=D\end{array}$

Substitute $-5\text{m}$ for $x\left(0\right)$ in above expression.

  $D=-5.0\text{m}$

Substitute $-5.0\text{m}$ for $D$ in equation (3).

  $\begin{array}{l}x\left(t\right)=\frac{\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^3}{3}+\left(9.5\text{m}/\text{s}\right)t+\left(-5.0\text{m}\right)\\ x\left(t\right)=\frac{\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^3}{3}+\left(9.5\text{m}/\text{s}\right)t-5.0\text{m}\end{array}$

Conclusion:

Thus, the position as function of time for the given interval is $x\left(t\right)=\frac{\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^3}{3}+\left(9.5\text{m}/\text{s}\right)t-5.0\text{m}$ .

(c)

To determine

The average velocity for the given time interval and compare to the average of instantaneous velocities of starting and ending times.

Answer to Problem 111P

The average velocity for the given time interval is $12.8\text{m}/\text{s}$ and it is not equal to the average of instantaneous velocities of the start and ending times.

Explanation of Solution

Given:

The acceleration of the particle is given by $a_x=\left(0.20\text{m}/{\text{s}}^{\text{3}}\right)t$ .

The velocity of particle at $t=0$ is $9.5\text{m}/\text{s}$ .

The position of particle at $t=0$ is $-5.0\text{m}$ .

Formula used:

Write expression for average velocity of the particle.

  $v_{av}=\frac{1}{\Delta t}{\displaystyle \underset{t=t_1}{\overset{t_2}{\int }}v\left(t\right)dt}$............. (4)

Write expression for instantaneous velocity of the particle.

  $v\left(t\right)=\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^2+9.5\text{m}/\text{s}$

  ........(5)

Write expression for average of instantaneous velocities for $t=0$ and $t=10$ .

  ${v^{\prime }}_{av}=\frac{v\left(10\right)+v\left(0\right)}{2}$............. (6)

Calculation:

Substitute $0\text{s}$ for $t_1$ , $10\text{s}$ for $t_2$ , $10\text{s}$ for $\Delta t$ and $\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^2+9.5\text{m}/\text{s}$ for $v\left(t\right)$ in equation (4).

  $\begin{array}{l}{v}_{av}=\frac{1}{10}{\displaystyle \underset{t=0}{\overset{10}{\int }}\left[\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^2+9.5\text{m}/\text{s}\right]dt}\\ v_{av}=\frac{1}{10}{\left(\frac{\left(0.10\text{m}/{\text{s}}^{\text{3}}\right)t^3}{3}+\left(9.5\text{m}/\text{s}\right)t\right)}_0^{10}\\ v_{av}=\frac{1}{10}\left[\frac{\left(0.10\text{m}/{\text{s}}^{\text{3}}\right){\left(10\right)}^3}{3}+\left(9.5\text{m}/\text{s}\right)\left(10\right)\right]\\ v_{av}=12.8\text{m}/\text{s}\end{array}$

Substitute $0\text{s}$ for $t$ in equation (5).

  $\begin{array}{l}v\left(0\right)=\left(0.10\text{m}/{\text{s}}^{\text{3}}\right){\left(0\right)}^2+9.5\text{m}/\text{s}\\ v\left(0\right)=9.5\text{m}/\text{s}\end{array}$

Substitute $10\text{s}$ for $t$ in equation (5).

  $\begin{array}{l}v\left(10\right)=\left(0.10\text{m}/{\text{s}}^{\text{3}}\right){\left(10\text{s}\right)}^2+9.5\text{m}/\text{s}\\ v\left(10\right)=\left[10+9.5\right]\text{m}/\text{s}\\ v\left(10\right)=19.5\text{m}/\text{s}\end{array}$

Substitute $19.5\text{m}/\text{s}$ for $v\left(10\right)$ and $0\text{m}/\text{s}$ for $v\left(0\right)$ in equation (6).

  $\begin{array}{l}{v^{\prime }}_{av}=\frac{19.5\text{m}/\text{s}+9.5\text{m}/\text{s}}{2}\\ {v^{\prime }}_{av}=14.5\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the average velocity for the given time interval is $12.8\text{m}/\text{s}$ and it is not equal to the average of instantaneous velocities of the start and ending times.