Question

A bus traveling along a straight highway covers one-
third of the total distance between two places with a
velocity 20 km h-!. The remaining part of the dis-
tance was covered with a velocity of 30 km h-' for
the first half of the remaining time and with velocity
50 km h-1 for the next half of the time. Find the
average velocity of the bus for its whole journey.
​

Answer 1

Answer:

$\frac{900}{31}$≈ 29 km/h

Explanation:

Given :

A bus traveling along a straight highway covers one-third of the total distance between two places with a velocity 20 km/h.

The remaining part of the distance was covered with a velocity of 30 km/h for the first half of the remaining time and with velocity 50 km h-1 for the next half of the time.

Find the average velocity of the bus for its whole journey.

​Solution :

We know that,

$Average \ speed = \frac{Total \ distance \ traveled}{Total \ time \ taken}$

Let the total distance be ''d',

Let the total time taken be "T" ,

Where T = $t_1 + t_2 + t_3$,

$t_1$ = time taken to cover 1/3 of the distance with speed 20 km/h,.

$t_2$ = time taken to cover remaining 1/2 of the distance with speed 30 km/h,.

$t_3$ = time taken to cover remaining distance with speed 50 km/h,.

Hence,

$t_1 = \frac{Distance}{Speed} = \frac{\frac{1}{3}d }{20}$..

⇒ $t_1=\frac{d}{60}$

Distance traveled in ($t_2$) = Half of remaining distance,

Remaining distance = (Total distance - distance already traveled)

$t_2 = \frac{Distance}{Speed} = \frac{(1 -\frac{1}{3})(\frac{1}{2})d}{30}$

⇒ $\frac{(1 -\frac{1}{3})(\frac{1}{2})d}{30} = \frac{(\frac{2}{3} )(\frac{1}{2})d}{30} = \frac{\frac{1}{3}d}{30}= \frac{d}{90}$

Distance traveled in time ($t_3$) is the remaining distnce

= (Total distance - distance already traveled) = $d - \frac{1}{3} d -(1 -\frac{1}{3} )(\frac{1}2})d$

$t_3 =\frac{Distance}{Speed}\frac{(1-(\frac{1}{3} + (1-\frac{1}{3})(\frac{1}{2})))d}{50}$

⇒ $t_3 =\frac{(1-(\frac{1}{3}+(1-\frac{1}{3})(\frac{1}{2})))d}{50} =\frac{(1-(\frac{1}{3}+(\frac{2}{3})(\frac{1}{2})))d}{50} =\frac{(1-(\frac{1}{3}+(\frac{1}{3})))d}{50} =\frac{(1-\frac{2}{3})d }{50} =\frac{\frac{1}{3}d}{50} = \frac{d}{150}$

∴ $Average \ speed = \frac{Total \ distance \ traveled}{Total \ time \ taken}$

⇒ $Average \ speed = \frac{d}{T} = \frac{d}{t_1+t_2+t_3} = \frac{d}{(\frac{d}{60})+(\frac{d}{90})+(\frac{d}{150} )} = \frac{d}{d(\frac{1}{60}+ \frac{1}{90}+ \frac{1}{150} )} = \frac{1}{\frac{15}{900}+\frac{10}{900} + \frac{6}{900} } = \frac{1}{\frac{31}{900}}=\frac{900}{31}$≈ 29 km/h

Answer 2

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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