Chapter 2, Problem 104P

(a)

To determine

To find: The magnitude of displacement in meters for the shaded box.

Answer to Problem 104P

The magnitude of displacement for the shaded region is $1\text{m}$ .

Explanation of Solution

Given:

The initial velocity on the box is $4\text{m}/\text{s}$ .

The final velocity on the box is $5\text{m}/\text{s}$ .

The time interval covered in one box is $1\text{s}$ .

Formula used:

The area of the velocity-time graph gives the displacement of the body.

Write the expression for the area of the shaded region.

  $A=\left(v_f-v_i\right)\left(t_f-t_i\right)$.........(1)

Here, $A$ is the area of the shaded region, $v_f$ is the upper point on the graph on the y-axis, $v_i$ is the lower point on the graph on the y-axis, $t_f$ is the upper point on the graph on the x-axis and $t_i$ is the lower point on the graph on the x-axis.

Calculation:

Substitute $4$ for $v_i$ , $5$ for $v_f$ , $1$ for $t_i$ and $2$ for $t_f$ in equation (1).

  $\begin{array}{c}A=\left(5-4\right)\left(2-1\right)\\ =1\text{m}\end{array}$

Conclusion:

Thus, the magnitude of displacement for the shaded region is $1\text{m}$ .

(b)

To determine

To find:The displacement between the two given time intervals of $1\text{s}$ .

Answer to Problem 104P

The displacement of the particle for the given $1\text{s}$ intervals is $2.5\text{m}$ and $3.1\text{m}$ respectively.

Explanation of Solution

Given:

The first $1\text{s}$ interval starts at $t=1.0\text{s}$ .

The second $1\text{s}$ interval starts at $t=2.0\text{s}$ .

Formula used:

The area of one block of the graph is $1$ , the area under the curve for the first time interval is almost $1.25$ block and area under the curve for the second time interval is almost $3.1$ block.

Write the expression for the displacement.

  $s=n.A$.........(2)

Here, $s$ is the displacement of the particle and $n$ is the number of blocks under the curve.

Calculation:

Substitute $s_1$ for $s$ , $1.25$ for $n$ and $1\text{m}$ for $A$ in equation (2).

  $\begin{array}{c}{s}_1=\left(1.25\right)\left(1\right)\text{m}\\ \text{=1}\text{.25}\text{m}\end{array}$

Here, $s_1$ is the displacement of the first interval.

Substitute $s_2$ for $s$ , $3.1$ for $n$ and $1\text{m}$ for $A$ in equation (2).

  $\begin{array}{c}{s}_2=\left(3.1\right)\left(1\right)\text{m}\\ \text{=3}\text{.1}\text{m}\end{array}$

Here, $s_2$ is the displacement of the second interval.

Conclusion:

Thus, the displacement of the particle for the given $1\text{s}$ intervals is $2.5\text{m}$ and $3.1\text{m}$ respectively.

(c)

To determine

To find:The average velocity of the particle for given time interval.

Answer to Problem 104P

The average velocity of the particle for the given time period is $2.17\text{m}/\text{s}$ .

Explanation of Solution

Given:

The distance covered in the first interval is $1.25\text{m}$ .

The distance covered in the second interval is $3.1\text{m}$ .

Formula used:

The area of one block of the graph is $1$ , the area under the curve for the given time interval is almost 4.35 blocks.

The average velocity is defined as the total displacement in the given time period.

Write expression for total displacement of the particle.

  $S=s_1+s_2$

…... (3)

Here, $S$ is the total displacement of particle.

Write the expression for the average velocity of the particle.

  $v_{av}=\frac{S}{T}$.........(4)

Here, $v_{av}$ is the average velocity of the particle and $T$ is the total time taken.

Calculation:

Substitute $1.25\text{m}$ for $s_1$ and $3.1\text{m}$ for $s_2$ in equation (3).

  $\begin{array}{c}S=1.25+3.1\\ =4.35\text{m}\end{array}$

Substitute $4.35\text{m}$ for $S$ and $2\text{s}$ for $T$ in equation (4).

  $\begin{array}{c}{v}_{av}=\frac{4.35}{2}\\ =2.17\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the average velocity of the particle for the given time period is $2.17\text{m}/\text{s}$ .

(d)

To determine

To explain: The average displacement of the particle for the given time period and compare it with the answer of part (b) and to compare the average velocity to the mean of initial and final velocity of the particle.

Answer to Problem 104P

The displacement of the particle for the interval $1.0\text{s}\le \text{t}\le \text{3}\text{.0}\text{s}$ is $4.3\text{m}$ , the average velocity is not equal to the mean of initial and final velocity for a given time interval.

Explanation of Solution

Given:

The equation of curve is $v_x=\left(0.50\text{m}/{\text{s}}^{\text{3}}\right)t^2$ .

The time interval is $1.0\text{s}\le \text{t}\le \text{3}\text{.0}\text{s}$ .

Formula used:

Write the expression for the velocity of particle.

  $v_x=\frac{dx}{dt}$

Rearrange the above expression for $dx$ .

  $dx=v_{x}dt$

Integrate the above expression from $x=0$ to $x=x$ and $t=1.0$ to $t=3.0$ .

  ${\displaystyle \underset{x=0}{\overset{x}{\int }}dx}={\displaystyle \underset{t=1.0}{\overset{3.0}{\int }}{v}_{x}dt}$.........(5)

Write an expression for the mean of initial and final velocity.

  $v_{mean}=\frac{v_{final}+v_{initial}}{2}$.........(6)

Calculation:

Substitute $\left(0.50\text{m}/{\text{s}}^{\text{3}}\right)t^2$ for $v_x$ in equation (5) and solve.

  $\begin{array}{c}x=\left(0.50\text{m}/{\text{s}}^{\text{3}}\right){\displaystyle \underset{1.0}{\overset{3.0}{\int }}{t}^{2}dt}\\ =\left(0.50\text{m}/{\text{s}}^{\text{3}}\right){\left[\frac{t^3}{3}\right]}_{1.0}^{3.0}\\ =4.3\text{m}\end{array}$

Substitute $0.5\text{m}/\text{s}$ for $v_{initial}$ and $4.25\text{m}/\text{s}$ for $v_{final}$ in equation (6).

  $\begin{array}{c}{v}_{mean}=\frac{4.25+0.5}{2}\\ =2.37\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the displacement of the particle for the interval $1.0\text{s}\le \text{t}\le \text{3}\text{.0}\text{s}$ is $4.3\text{m}$ , the average velocity is not equal to the mean of initial and final velocity for a given time interval.