Q11PE - College Physics (Urone)

Question

A student drove to the university from her home and noted that the odometer reading of her car increased by $\mathbf{12}\mathbf{km}$ . The trip took  $\mathbf{18}\mathbf{min}$. 

(a) What was her average speed? 

(b) If the straight-line distance from her home to the university is $\mathbf{10}.\mathbf{3}\mathbf{km}$ in a direction $\mathbf{25}.\mathbf{0}º$ south of east, what was her average velocity? 

(c) If she returned home by the same path $\mathbf{7}\mathbf{h}\mathbf{30}\mathbf{min}$ after she left, what were her average speed and velocity for the entire trip?

Step-by-Step Solution

Answer

1. $\mathbf{40}\mathbf{\text{\hspace{0.17em}km/h}}$
   
2. $\mathbf{34}\mathbf{.}\mathbf{33}\mathbf{\text{\hspace{0.17em}km/h }}$
   
3. $\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{\text{\hspace{0.17em}km/h}}$ and  $\mathbf{0}$

Step 1: Given Data

• Distance D $=\mathbf{12}\mathbf{km}$ .
• Time =  $\mathbf{18}\mathbf{min}\mathbf{=}\mathbf{\text{18/60\hspace{0.17em}h}}\mathbf{=}\mathbf{\text{0.3\hspace{0.17em}hour}}$

Step 2: Average Speed of the girl

a)

The average speed of the girl is the ratio of the distance covered to that of the time period.

In this case, the formula is :

$\mathbf{\text{Average\hspace{0.17em}Speed}}\mathbf{=}\frac{\mathbf{\text{Distance}}}{\mathbf{\text{Time}}}$

Substituting the values in the above expression, we get:

 $\begin{array}{c}\mathrm{s}=\frac{12\text{\hspace{0.17em}km}}{0.3\text{\hspace{0.17em}h}}\\ =40\text{\hspace{0.17em}km/h}\end{array}$

Hence the average speed when she goes to school is $\mathbf{40}\mathbf{km}\mathbf{/}\mathbf{hr}$ .

Step 3: Average Velocity of the girl

b)

The displacement to that of the time taken to cover that displacement is called the time period.

The displacement is the shortest length between the initial and final positions. 

The shortest distance from her school to her house is given in the question as $\mathbf{10}.\mathbf{3}\mathbf{km}$ .

Hence the displacement of the girl will be  $\mathbf{10}.\mathbf{3}\mathbf{km}$ .

$\mathrm{Average}\mathrm{Velocity}=\frac{\mathrm{Displacement}}{\mathrm{Time}}$

Substituting the values in the above expression, we get:

$\begin{array}{c}\mathrm{v}=\frac{10.3\text{\hspace{0.17em}km}}{0.3\text{\hspace{0.17em}h}}\\ =34.33\text{\hspace{0.17em}km/h}\end{array}$

The average velocity of the girl will be  $\mathbf{34}.\mathbf{33}\text{ k}\mathbf{m}/\mathbf{h}$.

Step 3: Average velocity and speed of the girl for time 7 h 30 min

c)

The total time is taken to cover the distance is  $\mathbf{7}\mathbf{h}\mathbf{30}\mathbf{min}$.

Hence it will be $\mathbf{7}.\mathbf{5}\mathbf{h}$ .

The girl is traveling a total of  $\mathbf{12}\mathbf{km}$ from home to school and coming back $\mathbf{12}\mathbf{km}$  from school to home.

Hence the total distance traveled  $=12\text{\hspace{0.17em}km}+12\text{\hspace{0.17em}km }=24\text{\hspace{0.17em}km}$ .

Average speed  can be calculated as:

$\mathbf{Average}\mathbf{Speed}\mathbf{=}\frac{\mathbf{Distance}}{\mathbf{Time}}$

 

Substituting the values in the above expression, we get:

$\begin{array}{c}\mathrm{s}=\frac{24\text{\hspace{0.17em}km}}{7.5\text{\hspace{0.17em}h}}\\ =3.2\text{\hspace{0.17em}km/h}\end{array}$

Hence the speed of the girl going to school and then returning is  $\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{\text{\hspace{0.17em}km/h}}$.

Here as the initial and the final position of the girl is her house, the minimum or shortest distance between the initial and final position will be zero.

Hence the displacement will be zero.

Therefore, the average velocity of the body will be zero