Time Speed and Distance Questions Answers

Up rate = 13 + 5 = 2.6 km per hour

Down rate = 28 + 5 = 5.6 km per hour

$therefore$ Speed of the current $= frac{1}{2} (5.6 - 2.6)$
=1.5 km p.h.

Walking $frac{3}{4}$ of original speed means taking $frac{4}{3}$ of original time. Let the original time be t.

$therefore frac{4}{3}t = t + 1frac{1}{2} text{ hrs.}$

$therefore frac{1}{3}t = 1frac{1}{2} text{ hrs. or } t = 3 times 1frac{1}{2} text{ , } i.e., 4frac{1}{2} text{ hrs.}$

Distance travelled

$= 10 Big(frac{2 times 21 times 24} {21+24} Big) = 224 text{ km.}$

Total length to be crossed by the train
= 110 + 132 = 242 m

$text{Time} = frac{242}{Big(36 times frac{5}{18}Big)} text {seconds} = 24.2 text{ seconds}$

In 20 minutes,  the truck covers the distance of$35 times frac{20}{60} text{ km, } = frac{35}{3} text{ km}$
while the tractor covers the distance of
  $20 times frac{20}{60} = frac{20}{3 } text{ km }$

$therefore frac{35}{3} = frac{20}{3} + x Rightarrow x = frac{15}{3} = 5$

Let the distance from office to home = y km then time taken for the round trip

$= frac{y}{x_1} + frac{y}{x_2} = frac{yleft(x_1 + x_2right)}{x_1 x_2}$

and total distance travelled in that round trip = $2y$ km

$therefore text{ Average speed } = 2y div frac{yleft(x_1 + x_2right)}{x_1x_2}$

$= frac{2y x_1x_2}{yleft(x_1 + x_2right)} = frac{2x_1x_2}{x_1 + x_2}$

Length of the bridge $= frac{36 times 1000}{3600} times 140 = 1400 text{m}$

$therefore 1.4 text{ km}$

Let the length of train be $l$ metres.

Speed be $x$ metre/sec

According to question,

$begin{aligned} & frac{l}{x} = 1.75 Rightarrow l = 1.75 x text{ m } quad ...(i) & text{ and } frac{1}{(x - 10)} = 1.5 Rightarrow l = 1.5 (x - 10) quad ...(ii) & text{From } (i) text{ and } (ii) & 1.75x = 1.5x - 15 & 1.75x - 1.5x = - 15 & therefore quad .25x = - 15 & x = frac{15 times 100}{125} = 60 & therefore quad l = 1.75 times 60 = 105.00 & 105 text{ metres} end{aligned}$

Speed of the train $= frac{110}{3} text{ m/s}$

when it cross a railway platform 165 metres, the time taken

$= dfrac{165 + 110}{frac{110}{3}}$

$= dfrac{275 times 3}{110}$

= 7.5 Seconds

$Rightarrow$Let the first person be walking faster with speed $x$ km/h and second walking with speed $y$ km/h.

Case I Both walking in same directions.
 $therefore$ Distance travelled by first person in 9 h = 9$x$ km
and distance travelled by second person in 9 h = 9$y$ km

As both are 27 km apart
$therefore 9x - 9y = 27 x - y = 3$$6 + y = 9 Rightarrow y = 3 km/h$$x-y=3$ ...(i)

Case II Both walking in opposite directions.
 $therefore$ Distance travelled by first person in 3 h = 3$x$
and distance travelled by second person in 3 h = 3$y$
So, by condition,  $3x + 3y = 27 Rightarrow x + y = 9$  ...(ii)

On adding Eqs. (i) and (ii), we get
$Rightarrow 2x = 12 Rightarrow x = 6 km/h$
Put the value of $x$ in Eq. (ii), we get
$6 + y = 9 Rightarrow y = 3 km/h$
So, their speeds are 6 km/h and 3 km/h.

Here, $x = 2 , y = 5$ and $t = 15$ min
 $therefore$ Required time = $frac{x times t}{y - x} = frac{2 times 15}{5-2}$

$= frac{2 times 15}{3} = 2 times 5 = 10 text{ min}$

Let v₁ be the velocity of faster train and v₂ be the velocity of slower train.

Case I If they move in the same direction, then

$dfrac{110 +130}{v_1 - v_2} = 60s Rightarrow v_1 - v_2 = 4$  ...(i)

Case II If they move in opposite directions,

$dfrac{110 +130}{v_1 + v_2} = 3 Rightarrow v_1 - v_2 = 80$  ...(ii)

On adding Eqs. (i) and (ii), we get $2v_1 = 84 Rightarrow v_1$ = 42m/s Now, putting the value of v₁ in Eq. (ii), we get

$42 + v_2 = 80, v_2 = 38 text{ m/s } = 3 times 4 times dfrac{10}{60} = 2km$