Tag: time - calcuation of distance

Questions Related to time - calcuation of distance

Choose the correct answers from the alternative given :
In a row. 25 trees are planted at equal distance from each other. The distance between 1st and 25th tree is 30 m.
What is the distance between 3rd and 15th tree?

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. $8$ m

  2. $15$ m

  3. $16$ m

  4. $18$ m

Show answer
Correct Option: B
Explanation:

Let distance between two consecutive trees be x m.
So, distance between 1st and 25th tree = 24x m = $30$ m
Distance between 3rd and 15th = 12x m = $15$

A boat travels with a speed of 15 km/hr in still water. In a river flowing at 5 km/hr, the boat travels some distance downstream and then returns. The ratio of average speed to the speed in still water is

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. 8 : 3

  2. 3 : 8

  3. 8 : 9

  4. 9 : 8

Show answer
Correct Option: C
Explanation:

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, $\displaystyle \frac{\text{Average speed}}{\text{Speed in still water}} = \frac{40/3}{15} = \frac{40}{45} = \frac{8}{9}$.

Two goods train each 500 m long, are running in opposite direction on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one.

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. 12 sec

  2. 24 sec

  3. 48 sec

  4. 60 sec

Show answer
Correct Option: B
Explanation:

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 $\left( 500 \times \dfrac{6}{125} \right ) = 24 sec.$

Choose the correct answer from the alternatives given.
Two trains for Bhopal leave Delhi at 9 am and 8:30 am and travel at 90 km/hr and 80 km/hr respectively. How many kilometres from Delhi will the two trains be together?

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. 360 km

  2. 320 km

  3. 270 km

  4. 280 km

Show answer
Correct Option: A
Explanation:

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 minutes to reach the school at an average speed of 12 km/hour. If he wants to reach the school in 15 minutes, his average speed (in km/hour) will be

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. 16

  2. 17

  3. 18

  4. 19

Show answer
Correct Option: A
Explanation:

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


Let he takes $15\ min$ to reach the school at an average speed $x\ km/hr$.

Therefore,
$20\times 12=15\times x$
$x=16\ km/hr$

Hence, this is the answer.

After travelling for 30 minutes a train meets an accident, due to which it has to stop for 45 minutes. Due to the accident its speed is also reduced to 2/3 of its former value and the train reaches its destination 1 hour 30 minutes late. Had the accident occurred 60 km later, the train would have reached 30 minutes earlier. The length of the journey is

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. 90 km

  2. 120 km

  3. 150 km

  4. 180 km

Show answer
Correct Option: B
Explanation:

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
$\therefore x=120:km$