A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of \( 7.5 kmh^{-1}\). 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 the magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]

Asked by Abhisek | 1 year ago |  167

1 Answer

Solution :-

Distance to the market = 2.5 km = 2500 m

Speed of the man  walking to the market= 5 km/h 

= 5 x (\( \dfrac{ (5}{18}\)) = 1.388 m/s

Speed of the man walking when he returns back home = 7.5 km/h 

= 7.5  x (\( \dfrac{ (5}{18}\)) = 2.08 m/s

(a) Magnitude of the average speed is zero since the displacement is zero

 

(b)

(i) Time taken to reach the market = \( \dfrac{Distance}{Speed}\)

\( \dfrac{ 2500}{1.388}\) = 1800 seconds = 30 minutes

So, the average speed over 0 to 30 minutes is 5 km/h or  1.388 m/s

(ii) Time taken to reach back home = \( \dfrac{Distance}{Speed}\)

=\( \dfrac{ 2500}{2.08}\) = 1200 seconds = 20 minutes

So, the average speed is

Average Speed over a interval of 50 minutes

\( \dfrac{Distance\;Covered}{Time\; Taken}\)

\( \dfrac{ (2500 + 2500)}{3000 }\)

\( \dfrac{ 5000}{3000}\)\( \dfrac{5}{3}\) = 1.66 m/s = 6 km/h

(iii) Average speed over an interval of 0 – 40 minutes 

\( \dfrac{Distance\;Covered}{Time\; Taken}\)

\( \dfrac{ (2500 + 1250)}{2400 }\) = 1.5625 seconds = 5.6 km/h

Answered by Pragya Singh | 1 year ago

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