Question

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5km/h. Finding the
market closed, he instantly turns and walks back home with a speed of 7.5 km/h. Find the magnitude of
average velocity and average speed of the man over the interval of time :
(a) 0 to 30 minutes
(b) 0 to 50 minutes
(c) 0 to 40 minutes

Answer 1

Explanation:

HE TRAVELS 2.5 KM IN HALF HOUR i.e..,30 mintues.

As per the information is given to us.

The market is 2.5km away from his home and he walked at speed of 5km/hr.

So, AFTER 0-30 MINUTES TIME INTERVAL....

1. average speed of man=5km/60minutes=83.33m/minutes.
2. average velocity of the man is equal to the average speed.

So, AFTER 0-50 MINUTES TIME INTERVAL....

1. AVETAGE SPEED-2.5+2.5=5km/50minutes that's result into 100m/minutes.
2. AVERAGE VELOCITY=2.5 km-2.5km/ 50minutes...(because velocity=displacement/ time).

AVERAGE VELOCITY=0

So, AFTER 0-40MINUTES TIME INTERVAL.....

1. AVERAGE SPEED=2.5km+1.25km/40minutes that's result into 93.75m/minutes.

AVERAGE VELOCITY=2.5-1.25/40=31.25m/minutes.

AS MAN WALKED BACK TO HIS HOME THE DISPLACEMENT OF THE MAN REDUCED AS MUCH HE TRAVELLED.

I SUPPOSE YOU UNDERSTAND IT AND IT MAY HELP YOU....

Answer 2

$Average\: velocity = \dfrac{total\: distance}{total\:time}$

$Time\: taken\: to\: go\: to\: market$

$\Rightarrow \dfrac{distance}{speed}$

$\dfrac{2.5}{7.5} = \dfrac{1}{3} = 20 minutes$

$Time\: taken\: to\: return$

$\Rightarrow \dfrac{distance}{speed}$

$\Rightarrow\dfrac{2.5}{5} = \dfrac{1}{2} = 30 minutes$

$Distance \:traveled \:in \:10\: minutes$

$while\: returning \:back =$

$\dfrac{10}{60} × 7.5 = 1.25km/hr[tex]</p><p>[tex] ( Speed\: is \:constant\: at \:7.5 km/hr )$

$Hence,\: Distance\: traveled\: in \:40\: minutes$

$(30 mins\: towards \:the\: market\: and$

$10\: mins\: towards \:home)$ $=2.5+1.25=3.75km$

$So \:average \:speed \:in \:0\: to\: 40\: min</p><p>$

[tex[\dfrac{3.75km}{\dfrac{40}{60}hr}[/tex]

$\dfrac{45}{8}km$