Question

A man walks on a straight road from his home to a market 2.5 kilometre away with speed of 5 kilometre per hour. Finding the market close, he instantly turns and walks back home with a speed of 7.5 kilometre per hour. What is the (a) magnitude of average velocity, and (b) average speed of the the man over the the interval of time (i) 0 to 30 minutes (ii) 0 to 50 minute (iii) 0 to 40 min?

Answer 1

Answer:

(a) 0

(b) (i) 5

(ii) 6

(iii) 5.625

Explanation:

(a) Magnitude of average velocity is 0 because displacement is 0 bacause initial position and final position is same.

(b) (i) Initially he traveled at a speed 5Km in an hr. So in 30 min he travels 2.5 km that means he reaches market in 30 min.

$avg \: speed = \frac{distance}{time}$

$avg \: speed = \frac{2.5}{ \frac{1}{2} }$

$avg \: speed = 5$

(ii) After the first 30min, i.e,. after reaching the market he travels at a speed of 7.5 km in 60 min. So in 20 min he travels 2.5 km. That means he reaches home. So the total distance traveled in first 50min is 2.5+2.5=5km

$avg \: speed = \frac{5}{ \frac{50}{60} } = 6kmph$

(iii) After the first 30min, i.e,. after reaching the market he travels at a speed of 7.5 km in 60 min. So in 10 min he travels 1.25 km. So the total distance traveled in first 40min is 2.5+1.25=3.75km

$avg \: speed = \frac{3.75}{ \frac{40}{60} } = 5.625kmph$