Question

(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 $40min$?

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Solution

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Speed with which he goes to market $=5km/h=5360010_{3} =1825 m/s$

Speed with which he comes back $=7.5km/h=7.5×360010_{3} =3675 m/s$

(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:$52.5 =1/2h=30$ 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 :$7.5×31 =2.5km$

His average speed in 0 to 50 minutes: $V_{avg}=timedistancetraveled $

$=(50/60)2.5+2.5 =6km/h$

(iii)In 40-30=10 minutes he travels a distance of :$7.5×61 =1.25km$

$V_{avg}=(40/60)2.5+1.25 =5.625km/h$

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