Question

a person walks along straight road from his house to market 2.5 kms away with a speed of 5kmperhr turns back reaches his house with speed 7.5 kmperhr average speed of person during time interval 0 to 50 min in m per sec​

Answer 1

Answer:

Explanation:

d = distance

v = speed

t = time

Going to market

d = 2.5 km

v = 5 km/h

As we know

v =d/t ---------(eqn 1)

t = d/v = 2.5/5 = 0.5 hrs

Coming back home

d = 2.5 km

v = 7.5 km/h

Therefore again using (eqn 1)

t = d/v = 2.5/7.5 = 1/3 = 0.33 hrs

Therefore total distance travelled

D = 5 km

Total time taken

T = 0.5 +0.33 = 0.83 hrs

0.83hrs = 0.83*60 = 49.8 minutes

We can approx take 50 minutes

Average speed = Total distance travelled / Total time taken

As we have to find it in m/s

D = 5000m

T = 50*60 = 3000s

Avg. Speed = 5000/3000 = 1.67

Therefore avg. Speed = 1.67m/s