Chapter 3, Problem 43P

(a)

To determine

Average acceleration between $t=6\text{s}$ and $t=11\text{s}$.

Answer to Problem 43P

The average acceleration is $2{\text{m}/\text{s}}^2$.

Explanation of Solution

Write the equation to find the average acceleration.

    $a_{\text{av}}=\frac{v_{\text{f}}-v_{\text{i}}}{t_{\text{f}}-t_{\text{i}}}$

Here, $a_{\text{av}}$ is the average acceleration, $v_{\text{f}}$ is the final velocity, $v_{\text{i}}$ is the initial velocity , $t_{\text{f}}$ is the final time and $t_{\text{i}}$ is the initial time.

Conclusion:

Substitute $14\text{m}/\text{s}$ for $v_{\text{f}}$, $4\text{m}/\text{s}$ for $v_{\text{i}}$, $11\text{s}$ for $t_{\text{f}}$ and $6\text{s}$ for $t_{\text{i}}$ in the above equation to find $a_{\text{av}}$.

    $\begin{array}{c}{a}_{\text{av}}=\frac{14\text{m}/\text{s}-4\text{m}/\text{s}}{11\text{s}-6\text{s}}\\ =2{\text{m}/\text{s}}^2\end{array}$

Thus, the average acceleration is $2{\text{m}/\text{s}}^2$.

(b)

To determine

Average velocity between $t=6\text{s}$ and $t=11\text{s}$.

Answer to Problem 43P

The average velocity is $9\text{m}/\text{s}$.

Explanation of Solution

Write the equation to find the average velocity.

    $v_{\text{av}}=\frac{\Delta x}{\Delta t}$        (I)

Here, $v_{\text{av}}$ is the average velocity, $\Delta x$ is the total displacement and $\Delta t$ is the total time.

The area under velocity time graph will give the displacement. Divide the area under the curve between $t=6\text{s}$ and $t=11\text{s}$ into two segments. They are given below.

A rectangle with sides $t=6\text{s}\text{to}11\text{s}$ and $v=0\text{m}/\text{s}\text{to}4\text{m}/\text{s}$. Its area is 20 m.

A right-angled triangle with sides $t=6\text{s}\text{to}11\text{s}$ and $v=4\text{m}/\text{s}\text{to}14\text{m}/\text{s}$. Its area is 25 m.

So, the total displacement during $t=6\text{s}\text{to}11\text{s}$ is 45 m.

Conclusion:

Substitute $45\text{m}$ for $\Delta x$ and $5\text{s}$ for $\Delta t$ in the equation for $v_{\text{av}}$.

    $\begin{array}{c}{v}_{\text{av}}=\frac{45\text{m}}{5\text{s}}\\ =9\text{m}/\text{s}\end{array}$

Thus, the average velocity is $9\text{m}/\text{s}$.

(c)

To determine

Average velocity between $t=0\text{s}$ and $t=20\text{s}$.

Answer to Problem 43P

The average velocity is $9.75\text{m}/\text{s}$.

Explanation of Solution

Divide the area into three segments. They are given below.

A rectangle with sides $t=0\text{s}\text{to}\text{20}\text{s}$ and $v=0\text{m}/\text{s}\text{to}4\text{m}/\text{s}$. Its area is 80 m.

A right-angled triangle with sides $t=6\text{s}\text{to}11\text{s}$ and $v=4\text{m}/\text{s}\text{to}14\text{m}/\text{s}$. Its area is 25 m.

A rectangle with sides $t=11\text{s}\text{to}20\text{s}$ and $v=4\text{m}/\text{s}\text{to}14\text{m}/\text{s}$. Its area is 90 m.

So, the total displacement during $t=0\text{s}$ to $t=20\text{s}$ is 195 m.

Conclusion:

Substitute $195\text{m}$ for $\Delta x$ and $20\text{s}$ for $\Delta t$ in the equation for $v_{\text{av}}$.

    $\begin{array}{c}{v}_{\text{av}}=\frac{195\text{m}}{20\text{s}}\\ =9.75\text{m}/\text{s}\end{array}$

Thus, the average velocity is $9.75\text{m}/\text{s}$.

(d)

To determine

The distance travelled from $t=10\text{s}$ and $t=15\text{s}$.

Answer to Problem 43P

The distance travelled is 69 m.

Explanation of Solution

Area under the graph from $t=10\text{s}$ to $t=15\text{s}$ will give the distance travelled. Area can be easily calculated using the number of unit squares. There are 68 complete squares and two half squares in the considered time interval. So, the total distance travelled will be 69 m.

Thus, the distance travelled is 69 m.