Solutions
Concept:
If the person is sitting in train and observing train velocity, he will observe the train is at rest as the relative velocity of train w.r.t person is zero.
VTP = VT – VP = VT – VT = 0
If the person is outside the train and stationery, he will observe the actual speed of train.
VTP = VT – VP = VT – 0 = VT
If the person is sitting in another vehicle and observing the train, the observed train velocity will be different.
VTP = VT – VP ⇒ VT = VTP + VP
Hence, the vectors VT, VP and VTP constitute a triangle.
Calculation:
Given uT = 54 kmph = 15 m/s; uB = uP = 0; aT = 0.2 m/s2; aP = 0.4 m/s2;
At t = 25 sec,
VT = uT + aT t = 15 + 0.2 × 25 = 20 m/s;
VP = uP + aP t = 0 + 0.4 × 25 = 10 m/s;
Now both the velocities’ directions are different, let’s solve in each direction separately,
We will assume, East direction as positive x – axis and North direction as positive Y – axis.
VTX = 20 × cos 30° = 17.32 m/s;
VTY = 20 × sin 30° = 10 m/s;
VPX = 10 × cos 0° = 10 m/s
VPY = 10 × sin 0° = 0 m/s;
VTPX = VTX – VPX = 7.32 m/s;
VTPY = VTY – VPY = 10 – 0 = 10 m/s;
Since Both the east component and north component are positive, the direction of Relative velocity vector lies in 1st quadrant,
∴ Passenger observes the train moving in north east direction (Option 2)
For angle,
⇒ θ = 53.8°
As the angle is measured from East direction, so θ = 53.8° (Option 3)
This problem can also be solved using Velocity triangle approach.
Mistake points:
The angle calculated (θ) is the angle with East direction (+ve x-axis)
If the angle is measured from north direction, then the answer will be 90 - θ degrees.