Question

A man walks on a straight road from his home to a market$2.5km$ away at a speed of $5km{h^{ - 1}}$. Finding the market closed, he instantly turns and walks back home with a speed of $7.5km{h^{ - 1}}$. What is the:
(a) Magnitude of average velocity, and
(b) Average speed of the man over the interval of time
(i) $0$ to $30$ min,
(ii) $0$ to $50$ min and
(iii) $0$ to $40$ min?

Answer 1

We can easily solve the given question, if we know about the calculation of distance and displacement. Simply we need to apply the formula for the velocity and speed in order to conclude with the correct solution for the given question.

Complete step by step solution:
For finding the average velocity, we need to find the total displacement.
Here, in this case the displacement is zero. Therefore, the average velocity will also be zero.
Total distance covered by the man $d = 2.5 + 2.5 = 5km$
We know that, $s = \dfrac{d}{t}$………. (i)
So, first of all let us find the time taken to reach the market. Initial speed is given as, $5km{h^{ - 1}}$.
So, from equation (i), we can write
$\Rightarrow t = \dfrac{d}{s} = \dfrac{{2.5}}{5} = 0.5hr = 30\min$
Now, the average speed from $0$ to $30$ min will be,
$\Rightarrow {s_1} = \dfrac{{2.5}}{{0.5}} = 5km{h^{ - 1}}$
Now, the distance travelled in $30$ min is $2.5km$ for which his average speed is $5km{h^{ - 1}}$. For the next $20$ min, he walks with a speed of $7.5km{h^{ - 1}}$ while returning to his home.
So, the distance covered in $20$ min can be given as,
$\Rightarrow {d_1} = {s_2}{t_1} = 7.5 \times \dfrac{{20}}{{60}} = 2.5km$
Now, the distance covered in$50$ min is ${d_2} = 2.5 + 2.5 = 5km$
Therefore, the average speed during $0$ to $50$ min will be,
$\Rightarrow {s_3} = \dfrac{5}{{\dfrac{{50}}{{60}}}} = 6km{h^{ - 1}}$
Now, the distance travelled in $30$ min is $2.5km$ for which his average speed is $5km{h^{ - 1}}$. For the next $10$ min, he walks with a speed of $7.5km{h^{ - 1}}$ while returning to his home.
So, the distance covered in $10$ min can be given as,
$\Rightarrow {d_3} = {s_2}{t_2} = 7.5 \times \dfrac{{10}}{{60}} = 1.25km$
Now, the distance covered in $40$ min is ${d_4} = 2.5 + 1.25 = 3.75km$
Therefore, the average speed during $0$ to $40$ min will be,
$\Rightarrow {s_3} = \dfrac{{3.75}}{{\dfrac{{40}}{{60}}}} = 5.625km{h^{ - 1}}$

Note: Distance is the total path covered whereas displacement is the shortest distance covered between two points which is given by a straight line. So, a body can have zero velocity even when it is moving with some speed as velocity depends on displacement and speed depends on distance.