Question

A motorcyclist drives from place A to B with a uniform speed of 30 km h-1 and returns from place B to A with a uniform speed of 20 km h -1. Find the average speed. ​

Answer 1

Hey Buddy

Here's the answer

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#We know average speed is ratio of total distance traveled to the total time taken

=> Avg speed = ( D1 + D1 )/( t1 + t2 )

Some important things

=> Let distance between A and B be ( x Km ) , so distance between B to A will also be ( X km )

So , from A to B

Speed = 30 Km/h

Distance = D1 = x Km

time = t 1

=> we know , Speed = D/t

therefore,

=> t = D/speed

t1 = x / 30

# Form B to A

Speed = 20 Km/h

Distance = D2 = x Km

time = t 2

=> we know , Speed = D/t

therefore ,

=> t = D/speed

t1 = x / 20

NOW PUTTING ALL VALUES IN MAIN EQN

Avg speed = ( D1 + D1 )/( t1 + t2 )

Avg Speed = ( x + x ) /( x/30 + x/20 )

Avg speed = ( 2 x )/ ( 5 x / 60 )

Avg speed = 2 x / ( x / 12 )

Avg speed = 2 x × 12 / x

Avg speed = 24 km / h

Average speed is 24 km / h


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HOPE HELPED

PEACE :)

Answer 2

it is a short cut trick ! try not to use in exam

Avg Speed = 2 AB/ A + B

A = speed in first trip

B = speed in second trip

Avg Speed = 2.30.20 / 20+30

= 1200 / 50

= 120/5

= 24 km/h