Motion in One Dimension Problem - 2

Description:

Question

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km hr^-1 finding the market closed, he instantly turns and walks back home with a speed of 7.5 km hr^-1. What is the

a. Magnitude of average velocity and

b. Average speed of the man over the interval of time

i. 0 to 30 min

ii. 0 to 50 min

iii. 0 to 40 min

Solution

a. Magnitude of average velocity −

V_a = Total displacement/time

As the starting and ending point of the man is same,

Total displacement = 0

∴ V_a = 0/t = 0

b.

Time to reach Market = Distance/speed

t₁ = 2.5 km/5 km hr^-1 = 1/2Hr = 30 minutes

Now, Time to reach back home from the market is given by −

t₂ = 2.5 km/7.5 km hr^-1 = 1/3Hr = 20 minutes

i. 0 to 30 min

V_a(0-30) = Total distance/Total time

V_a(0-30) = 2.5 km/30/60hr

V_a(0-30) = 2.5 km × 2 h^-1/1 = 5 km h^-1

ii. 0 to 50 min

V_a(0-50) = Total distance/Total time

V_a(0-50) = 5 km/50/60hr

V_a(0-50) = 6 km h^-1

iii. 0 to 40 min

V_a(0-40) = Total distance/Total time

V_a(0-40) = 3.75 km/40/60hr

V_a(0-40) = 5.63 km h^-1