Distance between Vinay's and Ganesh house is $9 k m$ . Vinay has to attend Ganesh birthday party at 7 O clock. He started from his home at 6 O clock on his bycycle and covered a distance of $6 K m$ in $40$ minutes at that point he met 'chintu' and he spoke to him for $5$ minutes reached Ganesh birthday at 7 O clock with what speed did he cover the second part of the journey? Calculate his average speed for the entire journey?

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Solution

The speed with which Vijay covered the second part of journey
$= \frac{D i s t a n c e l e f t t o r e a c h G a n e s h^{'} s h o u s e}{T i m e l e f t}$
$= \frac{9 k m - 6 k m}{1 h o u r - 45 m i n} = \frac{3 k m}{1 / 4 h} = 12 k m p h$
Average speed $= \frac{T o t a l d i s t a n c e c o v e r e d}{T o t a l t i m e t a k e n} = \frac{9 k m}{1 h} = 9 k m p h$

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The speed with which Vijay covered the second part of journey =Distance left to reach Ganesh′s houseTime left=9 km−6 km1 hour−45 min=3 km1/4 h=12 kmphAverage speed =Total distancecoveredTotal time taken=9 km1 h=9 kmph