Question

Arunoday started walking towards his office at 9AM in the morning. He reached his office at 9:30AM. He started walking again to reach home at the same time. What is the total distance covered by Arunoday if the average speed of Arunoday is $4km/h$?

Answer 1

Hint: We are given time and speed. Using the formula for distance, which relates all three distances, speed, and time we can calculate the value of distance asked in the question on putting the respective values appropriately.

Formula Used:
From the equations of motion that we have learnt we know that the distance covered is equal to the product of the velocity and the time taken. Thus,
$d = s \times t$
In this mathematical equation, $d$ represents the distance covered by the body and $t$ represents the time taken to cover this distance. $s$ represents the speed or the velocity of the body.

Complete step by step answer:
From the information conveyed by the numerical problem it is clear that the speed with which Arunoday walks is $4km/h$ . Thus, we can write:
$s = 4km{h^{ - 1}}$
Now Arunoday starts his journey at 9AM and reaches the office at 9:30AM, thus it takes him $30\min $ to reach his office. We can denote this time taken as ${t_1}$ . Thus,
${t_1} = 30\min = 0.5h$
Now we can use this information and substitute it to obtain the distance travelled by Arunoday in order to reach his office from his home. We can denote this distance as ${d_1}$.
Thus,
${d_1} = s \times {t_1} = 4 \times 0.5 = 2km$
Now Arunoday covers the same distance to reach his home from his office. If we denote this distance as ${d_2}$ . Thus,
${d_2} = s \times {t_2} = 4 \times 0.5 = 2km$
Thus the total distance covered by Arunoday is equal to the sum of these two distances. Thus,
$d = {d_1} + {d_2} = 2 + 2 = 4km$

Thus, Arunoday travels a total of $4km$ everyday.

Note: Make sure you convert the values given into SI units before using them in the question, like in this question, we have converted the time taken by Arunoday to travel from his home to office and office to home.