Chapter 3, Problem 41P

(a)

To determine

The acceleration $a_x$ at $t=11\text{s}$.

Answer to Problem 41P

The acceleration $a_x$ at $t=11\text{s}$ is $\underset{\_}{-10\text{m}/{\text{s}}^2}$.

Explanation of Solution

Use the given graph and write the expression for the acceleration $a_x$.

    $a_x=\frac{\Delta v_x}{\Delta t}$        (I)

Here, $a_x$ is the acceleration, $\Delta v_x$ is the change in velocity, $\Delta t$ is the time interval.

Conclusion:

Substitute $-10\text{m}/\text{s}$ for $\Delta v_x$, $1\text{s}$ for $\Delta t$ in equation (I) to find $a_x$.

    $\begin{array}{c}{a}_x=\frac{-10\text{m}/\text{s}}{1\text{s}}\\ =-10\text{m}/{\text{s}}^2\end{array}$

Therefore, the acceleration $a_x$ at $t=11\text{s}$ is $\underset{\_}{-10\text{m}/{\text{s}}^2}$.

(b)

To determine

The acceleration $a_x$ at $t=3\text{s}$.

Answer to Problem 41P

The acceleration $a_x$ at $t=3\text{s}$ is $\underset{\_}{0\text{m}/{\text{s}}^{\text{2}}}$.

Explanation of Solution

Use equation (I) to solve for $a_x$.

Conclusion:

Substitute $0\text{m}/\text{s}$ for $\Delta v_x$, $1\text{s}$ for $\Delta t$ in equation (I) to find $a_x$.

    $\begin{array}{c}{a}_x=\frac{0\text{m}/\text{s}}{1\text{s}}\\ =0\text{m}/{\text{s}}^2\end{array}$

The object is moving with a constant velocity.

Therefore, the acceleration $a_x$ at $t=3\text{s}$ is $\underset{\_}{0\text{m}/{\text{s}}^{\text{2}}}$.

(c)

To determine

The graph of acceleration $a_x\left(t\right)$.

Answer to Problem 41P

Figure $1$ shows the graph of acceleration $a_x\left(t\right)$.

Explanation of Solution

Use the equation (I) to solve for the acceleration $a_x\left(t\right)$.

$t\left(\text{s}\right)$	$0$	$2$	$4$	$6$	$8$	$10$	$12$	$14$
$a_x\left(t\right)\left(\text{m}/{\text{s}}^{\text{2}}\right)$	$0$	$0$	$0$	$5$	$5$	$-10$	$-10$	$0$

Figure $1$ shows the graph of acceleration $a_x\left(t\right)$.

Conclusion:

Therefore, the figure $1$ shows the graph of acceleration $a_x\left(t\right)$.

(d)

To determine

The distance travelled by the object from time $t=12\text{s}$ to $t=14\text{s}$.

Answer to Problem 41P

The distance travelled by the object from time $t=12\text{s}$ to $t=14\text{s}$ is $\underset{\_}{5.0\text{m}}$.

Explanation of Solution

Write the expression for the distance travelled by the object from time $t=12\text{s}$ to $t=14\text{s}$.

    $\text{distance}=\text{area under the graph from }t=12\text{s to }t=14\text{s}$        (II)

Write the expression for the read under the graph from time $t=12\text{s}$ to $t=14\text{s}$.

    $\text{distance}=\text{area of triangle ABC}$        (III)

Write the expression for the area of the triangle.

    $\text{area}=\frac{1}{2}\left(\text{BC}\right)\left(\text{AB}\right)$        (IV)

Use equation (IV) in (III) to solve for the distance travelled.

    $\text{distance}=\frac{1}{2}\left(\text{BC}\right)\left(\text{AB}\right)$        (V)

Conclusion:

Substitute $1\text{s}$ for $\text{BC}$, $10\text{m}/\text{s}$ in equation (V) to find distance travelled by the object from time $t=12\text{s}$ to $t=14\text{s}$.

    $\begin{array}{c}\text{distance}=\frac{1}{2}\left(1\text{s}\right)\left(10\text{m}/{\text{s}}^{\text{2}}\right)\\ =5.0\text{m}\end{array}$

Therefore, the distance travelled by the object from time $t=12\text{s}$ to $t=14\text{s}$ is $\underset{\_}{5.0\text{m}}$.