Chapter 2, Problem 110P

(a)

To determine

The general position function $x\left(t\right)$ .

Answer to Problem 110P

The position function for the given condition is $x\left(t\right)=\frac{1}{6}b{t}^3$ .

Explanation of Solution

Given:

The acceleration of the rocket is given by $a=bt$ .

The initial condition is given as $x_0=0$ and $v_x=v_{0x}$ .

Formula Used:

Write the expression for the acceleration of the rocket.

  $\frac{dv}{dt}=a$

Here, $a$ is the acceleration of the rocket and $\frac{dv}{dt}$ is the rate of change of velocity.

  $dv=a\cdot dt$

Substitute $bt$ for $a$ and integrate the above expression to and solve for $v$ .

  ${\displaystyle \int dv={\displaystyle \int bt\cdot dt}}$

Simplify the above expression.

  $v=\frac{1}{2}b{t}^2+C$........ (1)

Write expression for position of the rocket.

  $\begin{array}{l}\frac{dx}{dt}=v\\ dx=vdt\end{array}$

Substitute $\frac{1}{2}b{t}^2+C$ for $v$ in above expression.

  $dx=\left(\frac{1}{2}b{t}^2+C\right)dt$

Integrate the above expression.

  ${\displaystyle \int dx}={\displaystyle \int \left(\frac{1}{2}b{t}^2+C\right)dt}$

Simplify the above expression.

  $x=\frac{1}{6}b{t}^3+Ct+D$........ (2)

Calculation:

Substitute $v_{0x}$ for $v$ and $0$ for $t$ in equation (1) and solve for $C$ .

  $\begin{array}{c}{v}_{0x}=\frac{1}{2}\left(b\right)\left(0\right)+C\\ C=v_{0x}\end{array}$

Initially, the velocity of the rocket is zero. Substitute $0$ for $v_{0x}$ in above expression.

  $C=0$

Substitute $x_0$ for $x$ , $0$ for $t$ and $0$ for $C$ in equation (2) and solve for $D$ .

  $\begin{array}{c}{x}_0=\frac{1}{6}\left(b\right){\left(0\right)}^3+\left(0\right)\left(0\right)+D\\ D=x_0\end{array}$

Initially, position of rocket is $0$ . Substitute $0$ for $x_0$ in above expression.

  $D=0$

Express $x$ as function of time $x\left(t\right)$ . Substitute $0$ for C and $0$ for $D$ in equation (2).

  $x\left(t\right)=\frac{1}{6}b{t}^3+\left(0\right)t+\left(0\right)$

  $x\left(t\right)=\frac{1}{6}b{t}^3$........ (3)

Conclusion:

Thus, the position function for the given condition is $x\left(t\right)=\frac{1}{6}b{t}^3$ .

(b)

To determine

The position and velocity of the rocket for the given conditions.

Answer to Problem 110P

The position is $62.5\text{m}$ and velocity is $37.5\text{m}/\text{s}$ for the given condition.

Explanation of Solution

Given:

The initial position or rocket is $x_0=0$

The initial velocity of the rocket is $v_{0x}=0$

The time is $t=5\text{s}$ .

The value of $b$ is $3.0\text{m}/{\text{s}}^3$ .

Formula Used:

Write the expression of position function $x\left(t\right)$ .

  $x\left(t\right)=\frac{1}{6}b{t}^3$........ (3)

Calculation:

Substitute $5$ for $t$ and $3.0\text{m}/{\text{s}}^3$ for $b$ in equation (3).

  $\begin{array}{c}x\left(5\right)=\frac{1}{6}\left(3.0\text{m}/{\text{s}}^{\text{2}}\right){\left(5\text{s}\right)}^3\\ x\left(5\right)=62.5\text{m}\end{array}$

Substitute $5\text{s}$ for $t$ , $3.0\text{m}/{\text{s}}^3$ for $b$ and $0$ for $C$ in equation (1).

  $\begin{array}{l}v\left(5\right)=\frac{1}{2}\left(3.0{\text{m}/\text{s}}^3\right){\left(5\text{s}\right)}^2\\ v\left(5\right)=37.5\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the position is $62.5\text{m}$ and velocity is $37.5\text{m}/\text{s}$ for the given condition.

(c)

To determine

The average velocity for the given conditions and compare it with instantaneous velocity.

Answer to Problem 110P

The average velocity for the given condition is $37.6\text{m}/\text{s}$ . It is almost equal to the instantaneous velocity at $t=5$ .

Explanation of Solution

Given:

The time period is $t=4.5\text{s}$ to $t=5.5\text{s}$ .

Concept used:

Write expression for average velocity of the rocket.

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

Write expression for instantaneous velocity of the rocket.

  $v\left(t\right)=\frac{1}{2}b{t}^2$........ (5)

Calculation:

Substitute, $4.5\text{s}$ for $t_1$ and $5.5\text{s}$ for $t_2$ , $1\text{s}$ for $\Delta t$ and $\frac{1}{2}b{t}^2$ for $v\left(t\right)$ in equation (4).

  $\begin{array}{c}{v}_{av}=\frac{1}{1}{\displaystyle \underset{t=4.5}{\overset{5.5}{\int }}\frac{1}{2}b{t}^{2}dt}\\ v_{av}=\left[\frac{1}{6}b{\left(t^3\right)}_{4.5}^{5.5}\right]\\ =\left[\frac{1}{6}b\left\{{\left(5.5\right)}^3-{\left(4.5\right)}^3\right\}\right]\end{array}$

Substitute $3.0\text{m}/{\text{s}}^3$ for $b$ in above expression.

  $\begin{array}{c}{v}_{av}=\left[\frac{1}{6}\left(3\right)\left(75.25\right)\right]\text{m}/\text{s}\\ \text{=37}\text{.6}\text{m}/\text{s}\end{array}$

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

  $\begin{array}{c}v\left(5\right)=\frac{1}{2}\left(3\text{m}/{\text{s}}^3\right){\left(5\text{s}\right)}^2\\ =37.5\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the average velocity for the given condition is $37.6\text{m}/\text{s}$ . It is almost equal to the instantaneous velocity at $t=5$ .