Question

.A person travels along a straight road first half of distance with a speed 36kmph and second half of distance with a speed 72km/h find the average speed of person.

Answer 1

Let us assume that the distance covered by the man is 'd' km.

A person travels along a straight road first half of distance with a speed 36 km/hr.

Speed = 36 km/hr and distance = d km

Time = Distance/Speed

t1 = d/36 hr

Similarly, second half of distance with a speed 72km/hr.

t2 = d/72 hr

We have to the average speed of the person.

Average speed is defined as the ratio of total distance covered with respect to total time taken.

Total distance covered by person = d + d = 2d km

Total time taken by person = t1 + t2

= d/36 + d/72

= d/36 (1/1 + 1/2)

= d/36 (3/2)

= 3d/72

Average speed = 2d/(3d/72)

= (2d × 72)/3d

= 48

Therefore, the average speed of the person is 48 km/hr.

Shortcut Method:

s1 = 36 km/hr and s2 = 72 km/hr

Average speed = (2 × s1 × s2)/(s1 + s2)

= (2 × 36 × 72)/(36 + 72)

= 5184/108

= 48 km/hr

Answer 2

$\huge {\bf {\underline {Question}}}$

A person travels along a straight road first half of distance with a speed 36kmph and second half of distance with a speed 72km/h find the average speed of person.

$\huge {\bf {\underline {Solution}}}$

Let's assume that the distance covered by man is 'd' km.

The man travels first half distance of road with speed of 36 km/hr.

Speed = 36 km/hr and distance = d km

$\large {\fbox {\pink {\rm {Time = \frac{Distance}{Speed}}}}}$

$\rm {t1 = \frac{d}{36}hr}$

Similarly, man travels second half distance of road with speed of 72 km/hr.

$\rm {t2 = \frac{d}{72}hr}$

Now , we have to find the average speed of the man.

$\implies {Total\:distance\: covered = d + d =2d}$

$\implies {Total\:time\:taken = t1 + t2 = \frac{d}{36} + \frac{d}{72} = \frac {d}{36}( 1 + \frac{1}{2}) = {\frac {d}{36}} \times {\frac {3}{2}} = \frac {3d}{72}}$

$\large{Average\:speed = \frac{2d}{\frac{3d}{72}} = \frac {2d \times 72}{3d} = 48\:km/hr}$

Therefore, the average speed of the man is 48km/hr.