Question
Question: A bus X is travelling with a speed of \(40{\text{ }}\dfrac{{km}}{{hr}}\) towards North. Another bus ...
A bus X is travelling with a speed of 40 hrkm towards North. Another bus Y is travelling with a speed of 30 hrkm towards East. The velocity of bus X with respect to bus Y is?
Solution
For the given question, a bus X is travelling with a speed of 40 hrkm and another bus Y is travelling with a speed of 30 hrkm towards East, we have to find the velocity of X with respect to Y by vector addition. Then we will find the direction of the velocity from the trigonometric formula.
Complete answer:
It is given in the question that a bus X is travelling with a speed of 40 hrkm and another bus Y is travelling with a speed of 30 hrkm towards East. We have to find the velocity of bus X with respect to bus Y.
This is a 2-D problem which must be done by a careful vector addition.
Let us consider that the motion of X would be relative to Y if the bus X as stationary. With respect to Y, bus X would seem to be travelling toward the west at 30 hrkm.
Now, we the actual speed of bus X is namely 40 hrkm to the north.
The velocity vectors are perpendicular to each other and hence by vector addition we will find the answer.
Let the velocity of bus X with respect to bus Y is vr and the velocity of bus X is given vX=30 hrkm and the velocity of Y is vY=40 hrkm as shown in the diagram, and using pythagoras theorem we get,
Now,
vr=(vX)2+(vY)2
Substituting the values we get,
vr=(30)2+(40)2=2500=50
Hence, we get the velocity vr=50 hrkm.
The direction of the given velocity is, θ=tan−1vXvY=tan−13040=53.1∘
Note:
It must be noted that it is a two-dimensional problem which must be solved carefully by using vector addition. We must be careful in the direction of the velocity since the angle 53.1∘ here must be measured clockwise from the direction of west.