Compound measures and motion - class-VIII

Suppose he move 4 Km downstream in x hours.Then 
Speed downstream=$\frac{4}{n}$Km\hr
Speed upstream=$\frac{3}{n}$Km\hr
$\therefore \frac{48}{$\frac{4}{n}$}+\frac{48}{\frac{3}{x}}=14$ 
$\frac{48x}{4}+\frac{48x}{3}=14$
Or x=$\frac{336}{168}=2$
So spead downstream=$\frac{4}{\frac{1}{2}}=8$Km\hr
Or speed upstream =$\frac{3}{\frac{1}{2}}=6$Km\hr
So rate of stream =$\frac{1}{2}(8-6)=\frac{2}{2}=1$Km\hr

By definition,

If a car travels a distance d at rate $r _1$  and returns the same distance at rate $r _2$ , then 
Average speed = $\dfrac{total distance}{total time} = \dfrac{2d}{d/r _1+ d/r _2}= \dfrac{2r _1r _2}{r _1+r _2}$; 
$\therefore  x = \dfrac{2.30.40}{70} = \dfrac{240}{7} ~ 34 km/hr$.

Using the formula, $Speed=\frac { Distance }{ time } $
Mohit speed=
${ S } _{ m }=\frac { 225 }{ 4.5 } =50km/h$
${ S } _{ r }=50+10=60km/h$
Distance=225km
$Time=\frac { Distance }{ Speed } =\frac { 225 }{ 60 } =3\frac { 3 }{ 4 } hr$
Answer (C) $3\frac { 3 }{ 4 } hr$

Time taken in downstream $= \displaystyle \frac{d}{15 + 5} = \frac{d}{20}$
Time taken in upstream $= \displaystyle \frac{d}{15 - 5} = \frac{d}{10}$
$\therefore$ Average speed $ = \displaystyle \frac{2d}{\displaystyle \frac{d}{20} + \frac{d}{10}} = \frac{40}{3} km/hr$
So, $\displaystyle \frac{\text{Average speed}}{\text{Speed in still water}} = \frac{40/3}{15} = \frac{40}{45} = \frac{8}{9}$.

Relative speed = (45 + 30) km/hr
$ = \left ( 75 \times \dfrac{5}{18} \right ) m/sec$
$= \left( \dfrac{125}{6} \right) m/sec.$
We have to find the time taken by the slower train to pass the Driver of the faster train and not the complete train.
So, distance covered = Length of the slower train.
Therefore, Distance covered = 500 m.
$\therefore$ Required time $\left( 500 \times \dfrac{6}{125} \right ) = 24 sec.$

Distance travelled by the train at 80 km/hr in 30 min. = 40 km. This distance of 40 km is to be covered with a relative speed of
(90 - 80) = 10 km/hr.
Time taken to cover this distance = $\frac{40}{10}$ = 4 hr.
Hence, the trains will be together after 4 $\times $ 90=360 km from Delhi.

A boy takes $20\ min$ to reach the school at an average speed $12\ km/hr$.

Let the length of Journey = x km, and
Speed of the train = v km/hr

$\displaystyle\therefore30:min.+45:min.+\frac{\displaystyle\left(x-\frac{v}{2}\right)}{\displaystyle\frac{2}{3}v}$

$\displaystyle=\frac{x}{v}+1:hour:30:min.$

$\displaystyle\implies\frac{1}{2}+\frac{3}{4}+\frac{\displaystyle\frac{2x-v}{2}}{\displaystyle\frac{2v}{3}}=\frac{x}{v}+\frac{3}{2}$

$\displaystyle\implies\frac{3(2\times-v)}{4v}=\frac{x}{v}+\frac{1}{4}$

$\displaystyle=\frac{4x+v}{4v}$

$2x=4v$ ....(i)

Again, it accident occurred 60 kms later,

$\displaystyle\therefore30:min.+\frac{60}{v}+45:min.+\frac{\displaystyle\left(x-60-\frac{v}{2}\right)}{\displaystyle\frac{2}{3}v}$

$\displaystyle=\frac{x}{v}+1:hour:30:min.-30:min.$

$\displaystyle\implies\frac{1}{2}+\frac{60}{v}+\frac{3}{4}+\frac{3(2x-120-v)}{4v}=\frac{x}{v}+1$

$\displaystyle\implies\frac{x}{v}-\frac{60}{v}-\frac{3(2x-120-v)}{4v}=\frac{5}{4}-1$

$\displaystyle\implies\frac{4x-240-6x+360+3v}{4v}=\frac{1}{4}$

$\implies2x=2v+120$ ....(ii)
Solving eq. (i) and eq. (ii), we get
$\therefore x=120:km$