Chapter 2, Problem 112P

(a)

To determine

The instantaneous velocity as a function of time.

Answer to Problem 112P

The instantaneous velocity as the function of time is given as $v_x=v_{0x}+a_{0x}t+\frac{1}{2}b{t}^2$ .

Explanation of Solution

Given:

The acceleration of the particle is $a_x=a_{0x}+bt$ .

The position of particle is $x=x_0$ and $v_x=v_{0x}$ at $t=0$ .

Formula used:

Write expression for the acceleration of the particle.

  $a_x=\frac{d{v}_x}{dt}$

Here, $a_x$ is the acceleration of the particle and $\frac{d{v}_x}{dt}$ is the rate of change of velocity.

Rearrange above expression for $dv$ .

  $dv=a\cdot dt$  ........(1)

Calculation:

Substitute $a_{0x}+bt$ for $a_x$ and integrate equation (1) from $v_{0x}$ to $v_x$ and $0\text{s}$ to $t\text{s}$ .

  $\begin{array}{c}{\displaystyle \underset{v_{0x}}{\overset{v_x}{\int }}dv}={\displaystyle \underset{0}{\overset{t}{\int }}\left(a_x+bt\right)dt}\\ {\left[v_x\right]}_{v_{0x}}^{v_x}={\left[a_{0x}t+\frac{1}{2}b{t}^2\right]}_0^t\\ v_x-v_{0x}=a_{0x}t+\frac{1}{2}b{t}^2\\ v_x=v_{0x}+a_{0x}t+\frac{1}{2}b{t}^2\end{array}$

Conclusion:

Thus, the instantaneous velocity as the function of time is given as $v_x=v_{0x}+a_{0x}t+\frac{1}{2}b{t}^2$ .

(b)

To determine

The position of particle as function of time.

Answer to Problem 112P

The position of particle as a function of time is given by $x=x_o+v_{0x}t+\frac{a_{0x}{t}^2}{2}+\frac{1}{6}b{t}^3$ .

Explanation of Solution

Given:

The acceleration of the particle is $a_x=a_{0x}+bt$ .

The position of particle is $x=x_0$ and $v_x=v_{0x}$ at $t=0$ .

Formula used:

Write expression for the instantaneous velocity of the particle.

  $v_x=v_{0x}+a_{0x}t+\frac{1}{2}b{t}^2$

Write expression for the velocity of the particle.

  $v_x=\frac{dx}{dt}$  ........(2)

Rearrange above expression for $dx$ .

  $dx=v_{x}dt$

Calculation:

Substitute $v_{0x}+a_{0x}t+\frac{1}{2}b{t}^2$ for $v_x$ in equation (2) and integrate for $x_0$ to $x$ and $0$ to $t$ .

  $\begin{array}{c}{\displaystyle \underset{x_0}{\overset{x}{\int }}dx}={\displaystyle \underset{0}{\overset{t}{\int }}\left[v_{0x}+a_{0x}t+\frac{1}{2}b{t}^2\right]}dt\\ {\left[x\right]}_{x_0}^x={\left[v_{0x}t+\frac{a_{0x}{t}^2}{2}+\frac{1}{6}b{t}^3\right]}_0^t\\ x-x_0=v_{0x}t+\frac{a_{0x}{t}^2}{2}+\frac{1}{6}b{t}^3\\ x=x_o+v_{0x}t+\frac{a_{0x}{t}^2}{2}+\frac{1}{6}b{t}^3\end{array}$

Conclusion:

Thus, the position of particle as a function of time is given by $x=x_o+v_{0x}t+\frac{a_{0x}{t}^2}{2}+\frac{1}{6}b{t}^3$ .

(c)

To determine

The average velocity for the time interval $0$ to $t$ .

Answer to Problem 112P

The average velocity for the given time interval is given by $v_{av}=v_{0x}+\frac{a_{0x}t}{2}+\frac{1}{6}b{t}^2$ .

Explanation of Solution

Given:

The acceleration of the particle is $a_x=a_{0x}+bt$ .

The position of particle is $x=x_0$ and $v_x=v_{0x}$ at $t=0$ .

Formula used:

Write expression for average velocity of the particle.

  $v_{av}=\frac{1}{\Delta t}{\displaystyle \underset{t=t_1}{\overset{t_2}{\int }}v\left(t\right)dt}$  ........(3)

Calculation:

Substitute $0$ for $t_1$ , $t$ for $t_2$ , $\left(t-0\right)$ for $\Delta t$ and $x_o+v_{0x}t+\frac{a_{0x}{t}^2}{2}+\frac{1}{6}b{t}^3$ for $v\left(t\right)$ .

  $\begin{array}{l}{v}_{av}=\frac{1}{\left(t-0\right)}{\displaystyle \underset{0}{\overset{t}{\int }}\left(v_{0x}+a_{0x}t+\frac{1}{2}b{t}^2\right)dt}\\ v_{av}=\frac{1}{t}{\left[v_{0x}t+\frac{a_{0x}{t}^2}{2}+\frac{1}{6}b{t}^3\right]}_0^t\\ v_{av}=v_{0x}+\frac{a_{0x}t}{2}+\frac{1}{6}b{t}^2\end{array}$

Conclusion:

Thus, the average velocity for the given time interval is given by $v_{av}=v_{0x}+\frac{a_{0x}t}{2}+\frac{1}{6}b{t}^2$ .

(d)

To determine

The average of initial and final velocity and compare with part (c)

Answer to Problem 112P

The average of the initial and final velocities is given by ${v^{\prime }}_{av}=v_{0x}+\frac{1}{2}{a}_{0x}t+\frac{1}{4}b{t}^2$ . This is not equal to the average velocity for the given time interval.

Explanation of Solution

Given:

The acceleration of the particle is $a_x=a_{0x}+bt$ .

The position of particle is $x=x_0$ and $v_x=v_{0x}$ at $t=0$ .

Formula used:

Write expression for the average instantaneous velocity of the particle.

  ${v^{\prime }}_{av}=\frac{v_{ox}+v_x}{2}$  ........(4)

Calculation:

Substitute $v_{0x}+a_{0x}t+\frac{1}{2}b{t}^2$ for $v_x$ in equation (4).

  $\begin{array}{l}{v^{\prime }}_{av}=\frac{v_{ox}+v_{0x}+a_{0x}t+\frac{1}{2}b{t}^2}{2}\\ {v^{\prime }}_{av}=\frac{2v_{0x}+a_{0x}t+\frac{1}{2}b{t}^2}{2}\\ {v^{\prime }}_{av}=v_{0x}+\frac{1}{2}{a}_{0x}t+\frac{1}{4}b{t}^2\end{array}$

Conclusion:

Thus, the average of the initial and final velocities is given by ${v^{\prime }}_{av}=v_{0x}+\frac{1}{2}{a}_{0x}t+\frac{1}{4}b{t}^2$ . This is not equal to the average velocity for the given time interval.