Chapter 3, Problem 18P

(a)

To determine

Time took by the barge to reach the bridge.

Answer to Problem 18P

The time can be $\underset{\_}{22.5\text{s}}$ or $\underset{\_}{4.9\text{s}}$.

Explanation of Solution

Write the equation of motion

    $S=ut+\frac{1}{2}a{t}^2$        (I)

Here $S$ is the distance covered $u$ is the initial velocity, $a$ is the distance covered and $t$ is the time

Substitute the corresponding values for $S$, $u$ and $a$ in (I) and rearrange to solve the quadratic equation

    $\begin{array}{c}40\text{m}=\left(10\text{m/s}\right)t+\frac{1}{2}\left(-0.73{\text{m/s}}^{\text{2}}\right)t^2\\ 0=\left(0.365{\text{m/s}}^{\text{2}}\right)t^2-\left(10\text{m/s}\right)t+40\text{m}\\ t=\frac{-(-10.0\text{m}\text{}\text{/}\text{}\text{s})\pm \sqrt{{\text{(10}\text{.0 m/s )}}^{\text{2}}-\text{4(0}\text{.365}\text{m}\text{}\text{/}\text{}{\text{s}}^{\text{2}}\text{) (40}\text{.0 m)}}}{\text{2(0}\text{.365}\text{m}\text{}\text{/}\text{}{\text{s}}^{\text{2}}\text{)}}\\ =\text{22}\text{.5 s or 4}\text{.9 s}\end{array}$

Both the values of time are physically possible.

Conclusion:

The time can be $\underset{\_}{22.5\text{s}}$ or $\underset{\_}{4.9\text{s}}$.

(b)

To determine

Final velocities for each time.

Answer to Problem 18P

For $t=4.9\text{s}$, $v=6.4\text{m}/{\text{s}}^{\text{2}}$

For $t=22.5\text{s}$, $v=-6.4\text{m}/{\text{s}}^{\text{2}}$

Explanation of Solution

Write the equation of motion

    $v=u+at$        (II)

Conclusion:

Substitute $4.9\text{s}$ for $t$, $-0.73{\text{m/s}}^{\text{2}}$ for $a$ and $10\text{m/s}$ for $u$ in (II)

    $\begin{array}{c}v=\left(10\text{m/s}\right)+\left(-0.73{\text{m/s}}^{\text{2}}\right)\left(4.9\text{s}\right)\\ =6.4\text{m}/\text{s}\end{array}$

Substitute $22.5\text{s}$$4.9\text{s}$ for $t$, $-0.73{\text{m/s}}^{\text{2}}$ for $a$ and $10\text{m/s}$ for $u$ in (II)

    $\begin{array}{c}v=\left(10\text{m/s}\right)+\left(-0.73{\text{m/s}}^{\text{2}}\right)\left(22.5\text{s}\right)\\ =-6.4\text{m}/\text{s}\end{array}$

For $t=4.9\text{s}$, $v=6.4\text{m}/{\text{s}}^{\text{2}}$

For $t=22.5\text{s}$, $v=-6.4\text{m}/{\text{s}}^{\text{2}}$

(c)

To determine

Sketch velocity versus time and position versus time graphs.

Answer to Problem 18P

The graphs have been drawn.

Explanation of Solution

The short time corresponds to the time it takes the barge to reach the bridge, its velocity is in the positive direction. The longer time corresponds to the total time it takes to reach the bridge.

Conclusion:

The graphs have been drawn.