A car travels 1-3 of the distance on a straight road with a velocity of 10 km-h- next one-third with a velocity of 20 km-h and the last one third with a velocity of 60 km-h- Then the average velocity of the car -in km-h- during the whole journey is20183015

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Toppr Admin · Accepted Answer

Vavg-Total-xA0-DistanceTotal-xA0-timeTotal Distance -Svavg-SS3-xD7-10-S3-xD7-20-S3-xD7-60vavg-18m-sec

A car travels $1 / 3$ of the distance on a straight road with a velocity of $10 k m / h$ , next one-third with a velocity of $20 k m / h$ and the last one third with a velocity of $60 k m / h$ . Then the average velocity of the car (in km/h) during the whole journey is
$20$
$18$
$30$
$15$

$v_{a v g} = \frac{T o t a l D i s t a n c e}{T o t a l t i m e}$

Total Distance $= S$

$v_{a v g} = \frac{S}{\frac{S}{3 \times 10} + \frac{S}{3 \times 20} + \frac{S}{3 \times 60}}$

$v_{a v g} = 18 m / s e c$

A car travels 1/3 of the distance on a straight road with a velocity of 10km/h, next one-third with a velocity of 20km/h and the last one third with a velocity of 60km/h. Then the average velocity of the car (in km/h) during the whole journey is20183015

vavg=Total DistanceTotal timeTotal Distance =Svavg=SS3×10+S3×20+S3×60vavg=18m/sec

A car travels $1 / 3$ of the distance on a straight road with a velocity of $10 k m / h$ , next one-third with a velocity of $20 k m / h$ and the last one third with a velocity of $60 k m / h$ . Then the average velocity of the car (in km/h) during the whole journey is
$20$
$18$
$30$
$15$

$v_{a v g} = \frac{T o t a l D i s t a n c e}{T o t a l t i m e}$

Total Distance $= S$

$v_{a v g} = \frac{S}{\frac{S}{3 \times 10} + \frac{S}{3 \times 20} + \frac{S}{3 \times 60}}$

$v_{a v g} = 18 m / s e c$