Question

When going east at $ 10{\text{ km/h}} $ a train moving with constant velocity appears to be exactly north-east. When my velocity is increased to $ 30{\text{ km/h}} $ east it appears to be moving north. What is the velocity of the train along the north (in Kmph) ?
(A) $ 30{\text{ km/h}} $
(B) $ 20{\text{ km/h}} $
(C) $ 50{\text{ km/h}} $
(D) $ 10{\text{ km/h}} $

Answer 1

Hint: First of all we will find the speed of the train along the north and the east direction. The resultant velocity of the two velocities can give the actual speed of the train. The speed of the observer can be calculated as the initial velocity of the observer and the resultant velocity of the train.

Complete step by step solution:
Let the velocity of the moving train with the constant velocity with the speed, $ {v_1} = 10km/h $ in the east direction.
Let the train appear to be moving with the speed of $ {v_2} $ along the north-east direction.
In the basic directions, all the four directions are north, south, east and west from the right angle between each other.
The intermediate distance can be defined as the north-east makes an angle of $ 45^\circ $ to both the direction of the north and east side.
Therefore, the speed of the train along the north direction is
 $ {V_n} = {v_1}\tan \dfrac{\pi }{4} $
Place the values in the above expression-
 $
  {V_n} = 10 \times 1 \\
  {V_n} = 10km/h \\
  $
The speed of the train along the east direction can be given by $ {V_e} = {v_1}\tan \dfrac{\pi }{4} $
 $
  {V_e} = 10 \times 1 \\
  {V_e} = 10km/h \\
  $
Now, the actual speed of the train in north-east direction can be given by
 $ {V_2} = \sqrt {{{({v_n})}^2} + {{({v_e})}^2}} $
Place the values in the above equation-
 $
  {V_2} = \sqrt {{{(10)}^2} + {{(10)}^2}} \\
  {V_2} = \sqrt {100 + 100} \\
  {V_2} = \sqrt {200} \\
  {V_2} = 10\sqrt 2 km/h \\
  $
The speed at which the train move to the north side to that the train appears to be moving exactly in the south-East direction to be “v”
 $ {v_2} = v\cos \dfrac{\pi }{4} $
Make the velocity “v” the subject-
 $ v = \dfrac{{{v_2}}}{{\cos \dfrac{\pi }{4}}} $
Place the values in the above expression
 $ v = \dfrac{{10\sqrt 2 }}{{\dfrac{1}{{\sqrt 2 }}}} $
Simplify –
 $
  v = 10 \times 2 \\
  v = 20km/h \\
  $
From the given multiple choices, the option (B) is the correct answer.

Note:
The intermediate distance can be defined as the north-east makes an angle of $ 45^\circ $ to both the direction of the north and east side. The resultant velocity can be obtained from the square root of the sum of the squares of the individual velocity.