Chapter 2, Problem 62AP

(a)

To determine

Acceleration of object from $0\text{s}$ to $4\text{s}$.

Answer to Problem 62AP

Acceleration of object from $0\text{s}$ to $4\text{s}$ is $0\text{m}/{\text{s}}^{\text{2}}$

Explanation of Solution

Write the equation for acceleration.

    `$a=\frac{\Delta v}{\Delta t}$

Here, $a$ is the acceleration, $\Delta v$ is the change in velocity, and $\Delta t$ is the time interval.

The straight line parallel to x-axis of velocity-time graph from $0\text{s}$ to $4\text{s}$ shows that the velocity is constant.

Conclusion:

Substitute $0\text{m}/\text{s}$ for $\Delta v$ in the above equation to find $a$.

    $\begin{array}{c}a=\frac{0\text{m}/\text{s}}{\Delta t}\\ =0{\text{m}/\text{s}}^2\end{array}$

Therefore, the acceleration of object from $0\text{s}$ to $4\text{s}$ is $0\text{m}/{\text{s}}^{\text{2}}$

(b)

To determine

Acceleration of object from $4\text{s}$ to $9\text{s}$.

Answer to Problem 62AP

Acceleration of object from $4\text{s}$ to $9\text{s}$ is $6.0\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Write the equation for acceleration.

    `$a=\frac{\Delta v}{\Delta t}$

Write the equation for $\Delta v$.

    $\Delta v=v_f-v_i$

Here, $v_f$ is the final velocity and $v_i$ is the initial velocity.

Write the equation for $\Delta t$.

    $\Delta t=t_f-t_i$

Here, $t_f$ is the final time and $t_i$ is the initial time.

Rewrite the equation for $a$ by substituting the above relations.

    $a=\frac{v_f-v_i}{t_f-t_i}$                                                                                                          (I)

Conclusion:

Substitute $18\text{m}/\text{s}$ for $v_f$, $-12\text{m}/\text{s}$ for $v_i$, $9\text{s}$ for $t_f$, and $4\text{s}$ for $t_i$ in the above equation to find $a$.

    $\begin{array}{c}a=\frac{18\text{m}/\text{s}-\left(-12\text{m}/\text{s}\right)}{9\text{s}-4\text{s}}\\ =6.0\text{m}/{\text{s}}^2\end{array}$

Therefore, the acceleration of object from $4\text{s}$ to $9\text{s}$ is $6.0\text{m}/{\text{s}}^{\text{2}}$.

(c)

To determine

Acceleration of object from $13.0\text{s}$ to $18.0\text{s}$.

Answer to Problem 62AP

Acceleration of object from $13.0\text{s}$ to $18.0\text{s}$ is $-3.6\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Substitute $0\text{m}/\text{s}$ for $v_f$, $18\text{m}/\text{s}$ for $v_i$, $18\text{s}$ for $t_f$, and $13\text{s}$ for $t_i$ in the above equation to find $a$.

    $\begin{array}{c}a=\frac{0\text{m}/\text{s}-18\text{m}/\text{s}}{18\text{s}-13\text{s}}\\ =-3.6\text{m}/{\text{s}}^2\end{array}$

Therefore, the acceleration of object from $13.0\text{s}$ to $18.0\text{s}$ is $-3.6\text{m}/{\text{s}}^{\text{2}}$.

(d)

To determine

The time interval(s) at which the object moves with lowest speed.

Answer to Problem 62AP

Object moves with lowest speed at $6\text{s and at }18\text{s}$.

Explanation of Solution

Speed will be either zero or positive only. In the graph , it can be seen that $v_x$ is zero at $6\text{s and at }18\text{s}$.

Therefore, the object moves with lowest speed at $6\text{s and at }18\text{s}$.

(e)

To determine

The time at which object is at maximum distance from $x=0$.

Answer to Problem 62AP

The time at which object is at maximum distance from $x=0$ at $t=18\text{s}$.

Explanation of Solution

Initially, the object will goes into negative coordinates. At $t=6\text{s}$, object changes its direction. Up to $t=18\text{s}$, the object continues to move in same direction. So it can be say that the object is at maximum distance at $t=18\text{s}$.

Therefore, the time at which object is at maximum distance from $x=0$ at $t=18\text{s}$.

(f)

To determine

The final position of object at $t=18\text{s}$.

Answer to Problem 62AP

The final position of object at $t=18\text{s}$ will be $x=84\text{m}$.

Explanation of Solution

The distance travelled from velocity-time graph can be found using the area under the graph. The whole area of graph can be divided to five parts.

Write the equation to find the position of object at $t=18\text{s}$.

    $\Delta x=a_1{b}_1+\frac{1}{2}{a}_1{b}_2+\frac{1}{2}{a}_2{b}_3+a_2{b}_4+\frac{1}{2}{a}_2{b}_5$

Here, $\Delta x$ is the final position of object at $t=18\text{s}$, $a_1\text{and}{a}_2$ are lengths in y-coordinates, and $b_1,b_2,b_3,b_4,b_5$ are the lengths along x-axis.

Conclusion:

Substitute $-12\text{m}/\text{s}$ for $a_1$,$18\text{m}/\text{s}$ for $a_2$ $4\text{s}$ for $b_1$, $2\text{s}$ for $b_2$, $3\text{s}$ for $b_3$, $4\text{s}$ for $b_4$, and $5\text{s}$ for $b_5$ in the above equation to find $\Delta x$.

    $\begin{array}{c}\Delta x=\left(-12\text{m}/\text{s}\right)\left(4\text{s}\right)+\frac{1}{2}\left(-12\text{m}/\text{s}\right)\left(2\text{s}\right)+\frac{1}{2}\left(18\text{m}/\text{s}\right)\left(3\text{s}\right)+\left(18\text{m}/\text{s}\right)\left(4\text{s}\right)+\frac{1}{2}\left(18\text{m}/\text{s}\right)\left(5\text{s}\right)\\ =-48\text{m}-\text{12}\text{m+27}\text{m+72}\text{m+45}\text{m}\\ \text{=84}\text{m}\end{array}$

Therefore, the final position of object at $t=18\text{s}$ will be $x=84\text{m}$.

(g)

To determine

The total distance travelled by object during the time from $x=0\text{s}$ to $t=18\text{s}$.

Answer to Problem 62AP

The total distance travelled by object during the time from $x=0\text{s}$ to $t=18\text{s}$ is $204\text{m}$.

Explanation of Solution

Write the equation to find the total distance travelled by object.

    $x=\left|a_1{b}_1\right|+\left|\frac{1}{2}{a}_1{b}_2\right|+\left|\frac{1}{2}{a}_2{b}_3\right|+\left|a_2{b}_4\right|+\left|\frac{1}{2}{a}_2{b}_5\right|$

Here, $x$ is the total distance travelled.

Conclusion:

Substitute $-12\text{m}/\text{s}$ for $a_1$,$18\text{m}/\text{s}$ for $a_2$ $4\text{s}$ for $b_1$, $2\text{s}$ for $b_2$, $3\text{s}$ for $b_3$, $4\text{s}$ for $b_4$, and $5\text{s}$ for $b_5$ in the above equation to find $\Delta x$.

    $\begin{array}{c}\Delta x=\left|\left(-12\text{m}/\text{s}\right)\left(4\text{s}\right)\right|+\left|\frac{1}{2}\left(-12\text{m}/\text{s}\right)\left(2\text{s}\right)\right|+\left|\frac{1}{2}\left(18\text{m}/\text{s}\right)\left(3\text{s}\right)\right|\\ +\left|\left(18\text{m}/\text{s}\right)\left(4\text{s}\right)\right|+\left|\frac{1}{2}\left(18\text{m}/\text{s}\right)\left(5\text{s}\right)\right|\\ =\left|-48\text{m}\right|+\left|-\text{12}\text{m}\right|\text{+}\left|\text{27}\text{m}\right|\text{+}\left|\text{72}\text{m}\right|\text{+}\left|\text{45}\text{m}\right|\\ \text{=204}\text{m}\end{array}$

Therefore, the total distance travelled by object during the time from $x=0\text{s}$ to $t=18\text{s}$ is $204\text{m}$.