Chapter 2, Problem 63P

(a)

To determine

The average acceleration of given particle.

Answer to Problem 63P

The average acceleration in interval AB is $3.3\text{m}/{\text{s}}^{\text{2}}$ , the average acceleration in interval BC is $0.0\text{m}/{\text{s}}^{\text{2}}$ , the average acceleration in interval CE is $-7.50\text{m}/{\text{s}}^{\text{2}}$ .

Explanation of Solution

Given:

The variation of velocity of particle with time is plotted.

Formula used:

Write the expression for average acceleration.

  $a_{avg}=\frac{v_f-v_i}{\Delta t}$ ........ (1)

Here, $a_{avg}$ is average acceleration of particle, $v_f$ is the final velocity, $v_i$ is the initial velocity, and $\Delta t$ is the time interval.

Calculation:

Substitute $15\text{m}/\text{s}$ for $v_f$ , $5\text{m}/\text{s}$ for $v_i$ and $3\text{s}$ for $\Delta t$ in equation (1)

  $\begin{array}{c}{a}_{avg}{}_{AB}=\frac{15-5}{3}\\ =3.3\text{m}/{\text{s}}^{\text{2}}\end{array}$ .

Substitute $15\text{m}/\text{s}$ for $v_f$ , $15\text{m}/\text{s}$ for $v_i$ and $3\text{s}$ for $\Delta t$ in equation (1)

  $\begin{array}{c}{a}_{avgBC}=\frac{15-15}{3}\\ =0.0\text{m}/{\text{s}}^{\text{2}}\end{array}$ .

Substitute $-15\text{m}/\text{s}$ for $v_f$ , $15\text{m}/\text{s}$ for $v_i$ and $3\text{s}$ for $\Delta t$ in equation (1)

  $\begin{array}{c}{a}_{avgCE}=\frac{-15-15}{4}\\ =-7.50\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the average acceleration in interval AB is $3.3\text{m}/{\text{s}}^{\text{2}}$ , the average acceleration in interval BC is $0.0\text{m}/{\text{s}}^{\text{2}}$ , the average acceleration in interval CE is $-7.50\text{m}/{\text{s}}^{\text{2}}$ .

(b)

To determine

The positionof particle after given time interval.

Answer to Problem 63P

The particle is located at distance $75\text{m}$ from staring point after $10\text{s}$ .

Explanation of Solution

Formula used:

The displacement of the particle from the velocity-time graph is given by the area under the curve of the plot for the particular time interval.

The area under the graph for the interval $0-3\text{s}$ is a trapezium.

Write the expression for the area under the curve for the first interval.

  $\Delta x_{A\to B}=\frac{1}{2}\left(v_0+v_3\right)\left(\Delta t_1\right)$ ........ (1)

Here, $\Delta x_{A\to B}$ is the displacement in first interval, $v_0$ is the velocity at $0\text{s}$ , $v_3$ is the velocity at $3\text{s}$ and $\Delta t_1$ is the first time interval.

The area under the graph for the interval $3-6\text{s}$ is rectangle.

Write the expression for the area under the curve for the second interval.

  $\Delta x_{B\to C}=v_3\left(\Delta t_2\right)$ ........ (2)

Here, $\Delta x_{B\to C}$ is the displacement in second interval, $v_3$ is the velocity at $3\text{s}$ and $\Delta t_2$ is the second time interval.

The area under the graph for the interval $6-8\text{s}$ is triangle.

Write the expression for the area under the curve for the third interval.

  $\Delta x_{C\to D}=\frac{1}{2}\left(v_6\left(\Delta t_3\right)\right)$ ........ (3)

Here, $\Delta x_{C\to D}$ is the displacement in third interval, $v_6$ is the velocity at $6\text{s}$ and $\Delta t_3$ is the third time interval.

The area under the graph for the interval $8-10\text{s}$ is triangle.

Write the expression for the area under the curve for the fourth interval.

  $\Delta x_{D\to E}=\frac{1}{2}\left(v_{10}\left(\Delta t_4\right)\right)$ ........ (4)

Here, $\Delta x_{D\to E}$ is the displacement in fourth interval, $v_{10}$ is the velocity at $10\text{s}$ and $\Delta t_4$ is the third time interval.

Write the expression for displacement.

  $\Delta x=\Delta x_{A\to B}+\Delta x_{B\to C}+\Delta x_{C\to D}+\Delta x_{D\to E}$ ........ (5)

Here, $\Delta x$ is displacement in given time interval.

Calculation:

Substitute $5\text{m}/\text{s}$ for $v_0$ , $15\text{m}/\text{s}$ for $v_3$ and $3\text{s}$ for $\Delta t_1$ in equation (1).

  $\begin{array}{c}\Delta x_{A\to B}=\frac{1}{2}\left(5+15\right)\left(3\right)\\ =30\text{m}/\text{s}\end{array}$ .

Substitute $15\text{m}/\text{s}$ for $v_3$ and $3\text{s}$ for $\Delta t_2$ in equation (2).

  $\begin{array}{c}\Delta x_{B\to C}=\left(15\cdot 3\right)\\ =45\text{m}/\text{s}\end{array}$

Substitute $15\text{m}/\text{s}$ for $v_6$ and $2\text{s}$ for $\Delta t_3$ in equation (3).

  $\begin{array}{c}\Delta x_{C\to D}=\frac{1}{2}\left(15\cdot 2\right)\\ =15\text{m}/\text{s}\end{array}$

Substitute $-15\text{m}/\text{s}$ for $v_{10}$ and $2\text{s}$ for $\Delta t_4$ in equation (4).

  $\begin{array}{c}\Delta x_{D\to E}=\frac{1}{2}\left(-15\cdot 2\right)\\ =-15\text{m}/\text{s}\end{array}$ .

Substitute $30\text{m}/\text{s}$ for $\Delta x_{A\to B}$ , $45\text{m}/\text{s}$ for $\Delta x_{B\to C}$ , $15\text{m}/\text{s}$ for $\Delta x_{C\to D}$ and $-15\text{m}/\text{s}$ for $\Delta x_{D\to E}$ in equation (5).

  $\begin{array}{c}\Delta x=30+45+15+\left(-15\right)\\ =75\text{m}\end{array}$

Conclusion:

Thus the particle is located at distance $75\text{m}$ from staring point after $10\text{s}$ .

(c)

To determine

The plot of displacement of particle as a function of time

Answer to Problem 63P

The plot of displacement time graph is shown ion figure 1.

Explanation of Solution

The time interval BC is with constant velocity so in displacement time graph interval of BC will be straight line

  

Conclusion:

Thus, the plot of displacement time graph is shown ion figure 1.