Question

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h¯¹. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h¯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 ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]

Answer 1

Given: The distance from home to market is 2.5 km, the speed of man going to market is 5  kmh -1 and the speed of man coming back from market to home is 7.5  kmh −1 .

The time taken by man to reach the market from home is given as,

t 1 = d v 1

Where, d is the distance from home to market and v 1 is the speed of man going to market.

By substituting the values in the above expression, we get

t 1 = 2.5 5 = 1 2  h =30 min

The time taken by man to reach home from the market is given as,

t 2 = d v 2

Where, v 2 is the speed of man coming back to home.

By substituting the values in the above expression, we get

t 2 = 2.5 7.5 = 1 3  h =20 min

The total time for the journey is given as,

t= t 1 + t 2

By substituting the values in the above expression, we get

t=30+20 =50 min

(i)

For time interval 0 to 30 min,

The displacement of man is 2.5 km and the distance travelled by man is 2.5 km.

(a)

The magnitude of average velocity is given as,

v av = displacement t 1

By substituting the values in the above expression, we get

v av = 2.5 km ( 30 min× 1 h 60 min ) =5  kmh −1

Hence, the magnitude of average velocity is 5  kmh −1 .

(b)

The magnitude of average speed is given as,

s av = distance t 1

By substituting the values in the above expression, we get

s av = 2.5 km ( 30 min× 1 h 60 min ) =5  kmh −1

Hence, the average speed is 5  kmh −1 .

(ii)

For time interval 0 to 50 min,

The displacement of man is 0 km and the distance travelled by man is ( ( 2.5 km+2.5 km )=5 km ).

(a)

The magnitude of average velocity is,

v av = displacement t

By substituting the values in the above expression, we get

v av = 0 km ( 50 min× 1 h 60 min ) =0  kmh −1

Hence, the magnitude of average velocity is 0  kmh −1 .

(b)

The magnitude of average speed is given as,

s av = distance t

By substituting the values in the above expression, we get

s av = 5 km ( 50 min× 1 h 60 min ) =6  kmh −1

Hence, the magnitude of average speed is 6  kmh −1 .

(i)

For time interval 0 to 40 min,

The displacement of man is ( ( 2.5− 2.5 2 ) km= 5 4 km ) and the distance travelled by man is ( ( 2.5+ 2.5 2 ) km= 15 4 km ).

(a)

The magnitude of average velocity is given as,

v av = displacement t

By substituting the values in the above expression, we get

v av = 5 4  km ( 40 min× 1 h 60 min ) =1.875  kmh −1

Hence, the magnitude of average velocity is 1.875  kmh −1 .

(b)

The magnitude of average speed is given as,

s av = distance t

By substituting the values in the above expression, we get

s av = 15 4  km ( 40 min× 1 h 60 min ) =5.625  kmh −1

Hence, the magnitude of average speed is 5.625  kmh −1 .