Chapter 2, Problem 37P

(a)

To determine

$a_{\text{av,x}}$ Between $t=6\text{s}$ and $t=11\text{s}$.

Answer to Problem 37P

The average acceleration is $2.0\text{m}/{\text{s}}^2$.

Explanation of Solution

Write the expression for the average acceleration.

$\begin{array}{c}{a}_{\text{av,x}}=\frac{\Delta v_x}{\Delta t}\\ =\frac{v_2-v_1}{t_2-t_1}\end{array}$

Here, $a_{\text{av,x}}$ is the average acceleration along $x$ axis, $\Delta v_x$ is the change in velocity, $\Delta t$ is the change in time interval.

Conclusion:

Substitute $14\text{m}/s$ for $v_1$ and $4\text{m}/\text{s}$ for $v_2$, $11\text{s}$ for $t_1$ and $6\text{s}$ for $t_2$ in the above equation to find $a_{\text{av,x}}$.

$\begin{array}{c}{a}_{\text{av,x}}=\frac{14\text{m}/\text{s}-4\text{m}/\text{s}}{11\text{s}-6\text{s}}\\ =2.0\text{m}/{\text{s}}^2\end{array}$

Therefore, the average acceleration is $2.0\text{m}/{\text{s}}^2$.

(b)

To determine

$v_{\text{av,x}}$ for the interval between $t=6\text{s}$ and $t=11\text{s}$.

Answer to Problem 37P

$v_{\text{av,x}}$ for the interval between $t=6\text{s}$ and $t=11\text{s}$ is $9.0\text{m}/\text{s}$

Explanation of Solution

Write the expression for the average acceleration.

$v_{\text{av,x}}=\frac{\left(v_f+v_i\right)}{2}$

Here, $v_{\text{av,x}}$ is the average velocity along $x$ direction, $v_f$ is the final velocity, $v_i$ is the initial velocity.

Conclusion:

Substitute $14\text{m}/\text{s}$ for $v_f$ and $4\text{m}/\text{s}$ for $v_i$ in the above equation to find $v_{\text{av,x}}$.

$\begin{array}{c}{v}_{\text{av,x}}=\frac{14\text{m}/\text{s}+4\text{m}/\text{s}}{2}\\ =9.0\text{m}/\text{s}\end{array}$

Therefore, $v_{\text{av,x}}$ for the interval between $t=6\text{s}$ and $t=11\text{s}$ is $9.0\text{m}/\text{s}$

(c)

To determine

$v_{\text{av,x}}$ for the interval between $t=0\text{s}$ and $t=20\text{s}$.

Answer to Problem 37P

$v_{\text{av,x}}$ for the interval between $t=0\text{s}$ and $t=20\text{s}$ is $9.8\text{m}/\text{s}$

Explanation of Solution

Write the expression for average velocity.

$v_{\text{av,x}}=\frac{\Delta x}{\Delta t}$

Here, $\Delta x$ is the area under the graph in the figure.

Each square represents an area of $\left(1.0\text{m}/\text{s}\right)\left(1.0\text{s}\right)=1.0\text{m}$ and there are 195 squares under the graph.

Conclusion:

Substitute $1.0\text{m}\times \left(195\right)$ for $\Delta x$ and $20.0\text{s}$ for $\Delta t$ in the above equation to find $v_{\text{av,x}}$.

$\begin{array}{c}{v}_{\text{av,x}}=\frac{195\left(1.0\text{m}\right)}{20.0\text{s}}\\ =9.8\text{m}/\text{s}\end{array}$

Therefore, $v_{\text{av,x}}$ for the interval between $t=0\text{s}$ and $t=20\text{s}$ is $9.8\text{m}/\text{s}$

(d)

To determine

The increase in the cars speed between $10\text{s}$ and $15\text{s}$.

Answer to Problem 37P

The increase in speed between $10\text{s}$ and $15\text{s}$ is $2.0\text{m}/\text{s}$.

Explanation of Solution

At $t=10\text{s}$ the corresponding $v_x$ is $12\text{m}/\text{s}$ and from the graph, $t=15\text{s}$ $v_x=14\text{m}/\text{s}$.

Write the expression for change in speed.

$\Delta v_x=v_2-v_1$

$\Delta v_x$ is the change in velocity.

Conclusion:

Substitute $14\text{m}/\text{s}$ for $v_2$ and $12\text{m}/\text{s}$ for $v_1$ in the above equation to find $\Delta v_x$.

$\begin{array}{c}\Delta v_x=14\text{m}/\text{s}-12\text{m}/\text{s}\\ =2.0\text{m}/\text{s}\end{array}$

Therefore, the increase in speed between $10\text{s}$ and $15\text{s}$ is $2.0\text{m}/\text{s}$.

(e)

To determine

The displacement of the car from the time $t=10\text{s}$ to $t=15\text{s}$.

Answer to Problem 37P

The area under the $v_x$ versus t graph represents the displacement of the car.

Explanation of Solution

The area of the velocity time graph give the displacement. Displacement is the product of the velocity and time. To find the area under the graph, the area of the one cross section multiplied by the number of sections will give the total area.

Conclusion:

The area under the $v_x$ versus t graph represents the displacement of the car.