a bus travelling along a straight path covered the one third distance at velocity of 20kmh^1 .The half of the remaining part at 30kmh^-1 and the next half at 50kmh^-1.Find the average velocity.
Answers
We know that,
Average speed = Total distance traveled/Total time taken
Let the total distance be ''d',
Let the total time taken be "T", Where T = t1+ t2+ t3,
t1 = time taken to cover 1/3 of the distance with speed 20 km/h,
t2 = time taken to cover remaining 1/2 of the distance with speed 30 km/h,
t3 = time taken to cover remaining distance with speed 50 km/h.
Hence,
t1= Distance/Speed= [(1/3)d]/20
t1=d/60
Distance traveled in (t2) = Half of remaining distance,
Remaining distance = (Total distance - distance already traveled)
t2 = Distance/Speed= {(1 -1/3)×(1/2)×d}/30
{(1-1/3)×(1/2)×d}/30
{(2/3)×(1×2)×d/30
{(1/3)×d/30}
d/90
Distance traveled in time (t3) is the remaining distance
= (Total distance - distance traveled)
= d-1/3×d -(1-1/3)×(1/2)×d
t3 =Distance/Speed
={1-(1/3)+(1-1/3)×(1/2)×d}/50
=t3 ={1-(1/3)+(1-1/3)×(1×2)×d}/50
={1-(1/3)+(2/3)×(1/2)×d}/50
={1-(1/3)+(1/3)×d}/50
={1-2/3×d}/50
={(1/3)×d}/50
= d/150
∴ Average/speed = Total distance traveled/Total time taken
Average /speed = d/T
=d/{t1+t2+t3}
=d{(d/60)+(d/90)+(d/150)}
=d/{d(1/60)+(1/90)+(1/150)}
= 1/{(15/900)+(10/900)+(6/900)}
= 1/(31/900)
=900/31
=29.0322... km/h
=29 km/h (approx.)
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