When going east at 10km/h a train moving with constant velocity appears to be moving exactly north-east, when my velocity is increased to $30km/h$ east it appears to be moving north.
At what speed (in kmph) should I move north so that the train appears to be moving exactly South-East?
A. $30km/h$
B. $20km/h$
C. $50km/h$
D. $10km/h$

Hint: We have to find the speed of the train along the north and east direction. The resultant velocity of these two velocities give the actual speed of the train. The speed of the observer is calculated from the initial velocity of the observer and the resultant velocity of the train.

Complete step by step answer:
When I am moving with constant velocity with a speed of ${v_1} = 10km/h$ in the east direction, a train appears to be moving with a speed of ${v_2}$ along the north-east direction. In a cardinal direction, all the directions i.e., north, south, east and west form a right angle between each other. The intermediate distance between north and east i.e., north-east direction makes an angle of 45⁰ with each of the north and east side.
Hence, the speed of the train along the north direction is $Vn = {v_1}\tan \dfrac{\pi }{4}$
$Vn = 10 \times 1\left[ {\tan \dfrac{\pi }{4} = 1} \right]$
$\implies Vn = 10km/h$
The speed of the train along the east direction is $Ve = {v_1}\tan \dfrac{\pi }{4}$
$Ve = 10 \times 1$
$\implies Ve = 10km/h$
Now, the actual speed of the train in north-east direction is ${v_2}$
${v_2} = \sqrt {{{\left( {Vn} \right)}^2} + {{\left( {Ve} \right)}^2}} $
$\implies {v_2} = \sqrt {{{\left( {10} \right)}^2} + {{\left( {10} \right)}^2}} $
$\implies {v_2} = \sqrt {100 + 100} $
$\implies {v_2} = \sqrt {200} i.e.,10\sqrt 2 km/h$
The speed at which I should move to the north side so that the train appears to be moving exactly in the South-East direction be v.
${v_2} = v\cos \dfrac{\pi }{4}$
$\implies v = \dfrac{{{v_2}}}{{\cos \dfrac{\pi }{4}}}$
$\implies v = \dfrac{{10\sqrt 2 }}{{\dfrac{1}{{\sqrt 2 }}}}\left[ {\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}} \right]$
$\implies v = 10\sqrt 2 \times \sqrt 2 $
$\implies v = 10 \times 2$
$\therefore v = 20km/h$.

So, the correct answer is “Option B”.

Note:
The intermediate distances i.e., northeast makes an angle of 45⁰ to both the direction of north and east side. The resultant velocity is obtained from the square root of the sum of the squares of the individual velocity.