A man runs $200 m$ along a straight track in the first $25 s$ and then turns back instantly and runs $100 m$ in next $15 s$ towards the straight point. The ratio of magnitude of average velocity and average speed of the man is:

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Solution

Average velocity= total displacement / total time taken
$v_{a} v = \frac{s}{t} = \frac{200 - 100}{25 + 15} = \frac{100}{40} = 2.5 m / s$
$S p e e d_{a} v = \frac{d i s t a n c e}{t i m e} = \frac{200 + 100}{40} = \frac{300}{40} = 7.5 m / s$
Ratio , $\frac{v_{a} v}{S p e e d_{a} v} = \frac{2.5}{7.5}$
$= \frac{1}{3}$

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A man runs 200 m along a straight track in the first 25 s and then turns back instantly and runs 100 m in next 15 s towards the straight point. The ratio of magnitude of average velocity and average speed of the man is:

Average velocity= total displacement / total time taken vav=st=200−10025+15=10040=2.5m/sSpeedav=distancetime=200+10040=30040=7.5m/sRatio , vavSpeedav=2.57.5 =13

A man runs $200 m$ along a straight track in the first $25 s$ and then turns back instantly and runs $100 m$ in next $15 s$ towards the straight point. The ratio of magnitude of average velocity and average speed of the man is: