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# 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?

Solution
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#### Distance to market s=2.5km=2.5×103=2500mSpeed with which he goes to market =5km/h=51033600=2518m/sSpeed with which he comes back =7.5km/h=7.5×1033600=7536m/s(a)Average velocity is zero since his displacement is zero.(b) (i)Since the initial speed is 5km/s and the market is 2.5 km away,time taken to reach market:2.55=1/2h=30 minutes.Average speed over this interval =5km/h(ii)After 30 minutes,the man is travelling wuth 7.5 km/h speed for 50-30=20 minutes.The distance he covers in 20 minutes :7.5×13=2.5kmHis average speed in 0 to 50 minutes: Vavg=distancetraveledtime=2.5+2.5(50/60)=6km/h(iii)In 40-30=10 minutes he travels a distance of :7.5×16=1.25kmVavg=2.5+1.25(40/60)=5.625km/h

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Similar Questions
Q1
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h1 . Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h1. 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?
View Solution
Q2
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5kmh1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5kmh1. What is the magnitude of average velocity?
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Q3

A man walks on a straight road from his home to a market 2.5km away with a speed of 5km/h. Finding the market closed, he instantly turns and walk back home with a speed of 7.5 km/h. The average speed of the man over the, the interval of time 0 to 40min is equal to?

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Q4
(i) Explain clearly, with examples, the distinction between :
(a) magnitude of displacement over an interval of time and the total length of path covered by a particle over the same interval;
(b) magnitude of average velocity over an interval of time, and the average speed over the same interval.
Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ?

(ii) A man walks on a straight road from his home to a market 2.5km away with a speed of 5kmh1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5kmh1. What is the:
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (i) 0 to 30min, (ii) 0 to 50min, (iii) 0 to 40min?
View Solution
Q5
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 !]
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