Chapter 3, Problem 68A

(a)

To determine

To Find: The magnitude of displacement in the given time interval.

Answer to Problem 68A

  $87.5\text{ m}$

Explanation of Solution

Given:

Time interval: 5 min to 10 min

  

Formula Used:

Area under the velocity-time graph gives the magnitude of displacement.

Area of trapezium:

  $A=\frac{1}{2}\left(a+b\right)h$

Where, a and b are lengths of parallel sides and h is the distance between them.

Calculations:

Refer to the given graph for time interval 5 min to 10 min.

  $\begin{array}{c}{A}_{5-10}=\frac{1}{2}\left(20+15\right)\left(10-5\right)\\ =87.5\text{m}\end{array}$

Conclusion:

Thus, the magnitude of displacementfor time interval 5 min to 10 min is $87.5\text{ m}$ .

(b)

To determine

To Find: The magnitude of displacement in the given time interval.

Answer to Problem 68A

  $75\text{ m}$

Explanation of Solution

Given:

Time interval: 10 min to 15 min

  

Formula Used:

Area under the velocity-time graph gives the magnitude of displacement.

Area of rectangle:

  $A=a\times b$

Where, a is the length andbis the width.

Calculations:

Refer to the given graph for time interval 10 min to 15 min.

  $\begin{array}{c}{A}_{10-15}=\left(15\right)\left(15-10\right)\\ =75\text{m}\end{array}$

Conclusion:

Thus, the magnitude of displacement for time interval 10 min to 15 min is $75\text{ m}$ .

(c)

To determine

To Find: The magnitude of displacement in the given time interval.

Answer to Problem 68A

  $12.5\text{m}$

Explanation of Solution

Given:

Time interval: 25 min to 30 min

  

Formula Used:

Area under the velocity-time graph gives the magnitude of displacement.

Area of triangle:

  $A=\frac{1}{2}b\times h$

Where, b is the base and h is the height.

Calculations:

Refer to the given graph for time interval 25 min to 30 min.

  $\begin{array}{c}{A}_{25-30}=\frac{1}{2}\left(30-25\right)\times 5\\ =12.5\text{m}\end{array}$

Conclusion:

Thus, the magnitude of displacement for time interval 25 min to 30 min is $12.5\text{m}$ .

(d)

To determine

To Find: The magnitude of displacement in the given time interval.

Answer to Problem 68A

  $287.5\text{m}$

Explanation of Solution

Given:

Time interval: 0 min to 25 min

  

Formula Used:

Area under the velocity-time graph gives the magnitude of displacement.

Area of triangle:

  $A=\frac{1}{2}b\times h$

Where, b is the base and h is the height.

Area of rectangle:

  $A_r=l\times w$

Where, l is the length and w is the width of the rectangle.

Area of trapezium:

  $A_t=\frac{1}{2}\left(x+y\right)d$

Where, x and y are lengths of parallel sides and d is the distance between them.

Calculations:

Refer to the given graph for time interval 0 min to 5 min.

  $\begin{array}{c}{A}_{0-5}=\frac{1}{2}\left(b\times h\right)\\ =\frac{1}{2}\left(5-0\right)\left(20\right)\\ =50\text{m}\end{array}$

From part (a)

  $A_{5-10}=87.5\text{m}$

From part (b)

  $A_{10-15}=75\text{m}$

Refer to the given graph for time interval 15 min to 20 min.

  $\begin{array}{c}{A}_{15-20}=\frac{1}{2}\left(15+5\right)\left(20-15\right)\\ =50\text{m}\end{array}$

Refer to the given graph for time interval 20 min to 25 min.

  $\begin{array}{c}{A}_{20-25}=l\times w\\ =5\times \left(25-20\right)\\ =25\text{m}\end{array}$

Total displacement for time interval 0 min to 25 min is:

  $\begin{array}{l}{s}_{0-25}=A_{0-5}+A_{5-10}+A_{10-15}+A_{15-20}+A_{20-25}\\ s_{0-25}=50\text{m + 87}\text{.5 m +}\text{75 m + 50 m + 25 m }\\ {\text{s}}_{0-25}=287.5\text{m}\end{array}$

Conclusion:

Thus, the magnitude of displacement for time interval 0 min to 25 min is $287.5\text{m}$ .