A man walks on a straight road from his home to market 2.5 km away with a speed of 5 kmph . Finding the market closed he instantly turns and walks back home with a speed of 7.5 kmph
(a)what is the magnitude of average velocity
(b)average speed on the man over the interval of time
(c)a)0 to 30 min b)0 to 40 min c)0 to 50 min
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As the distance remains same
So V avg = 2v1v2/v1 + v2
= 2*5*7.5/12.5 = 6kmph
Now t1 = 2.5/5 = 0.3 hrs = 30min
So V avg is 5 kmph
t2 = 2/7.5 = 20 min So after 10 min he would be at the halfway during his return journey
Total distance = 3.75km
So speed = 3.75*6/4= 5.625
After 50min he would reach back and hence avg speed is 6 kmph
So V avg = 2v1v2/v1 + v2
= 2*5*7.5/12.5 = 6kmph
Now t1 = 2.5/5 = 0.3 hrs = 30min
So V avg is 5 kmph
t2 = 2/7.5 = 20 min So after 10 min he would be at the halfway during his return journey
Total distance = 3.75km
So speed = 3.75*6/4= 5.625
After 50min he would reach back and hence avg speed is 6 kmph
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