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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 kmhr40{\text{ }}\dfrac{{km}}{{hr}} towards North. Another bus Y is travelling with a speed of 30 kmhr30{\text{ }}\dfrac{{km}}{{hr}} towards East. The velocity of bus X with respect to bus Y is?

Explanation

Solution

For the given question, a bus X is travelling with a speed of 40 kmhr40{\text{ }}\dfrac{{km}}{{hr}} and another bus Y is travelling with a speed of 30 kmhr30{\text{ }}\dfrac{{km}}{{hr}} 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 kmhr40{\text{ }}\dfrac{{km}}{{hr}} and another bus Y is travelling with a speed of 30 kmhr30{\text{ }}\dfrac{{km}}{{hr}} towards East. We have to find the velocity of bus X with respect to bus Y.
This is a 22-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 kmhr30{\text{ }}\dfrac{{km}}{{hr}}.
Now, we the actual speed of bus X is namely 40 kmhr40{\text{ }}\dfrac{{km}}{{hr}} 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{v_r} and the velocity of bus X is given vX=30 kmhr{v_X} = 30{\text{ }}\dfrac{{km}}{{hr}} and the velocity of Y is vY=40 kmhr{v_Y} = 40{\text{ }}\dfrac{{km}}{{hr}} as shown in the diagram, and using pythagoras theorem we get,

Now,
vr=(vX)2+(vY)2{v_r} = \sqrt {{{\left( {{v_{_X}}} \right)}^2} + {{\left( {{v_Y}} \right)}^2}}
Substituting the values we get,
vr=(30)2+(40)2=2500=50{v_r} = \sqrt {{{\left( {30} \right)}^2} + {{\left( {40} \right)}^2}} = \sqrt {2500} = 50
Hence, we get the velocity vr=50 kmhr{v_r} = 50{\text{ }}\dfrac{{km}}{{hr}}.
The direction of the given velocity is, θ=tan1vYvX=tan14030=53.1\theta = {\tan ^{ - 1}}\dfrac{{{v_Y}}}{{{v_X}}} = {\tan ^{ - 1}}\dfrac{{40}}{{30}} = {53.1^ \circ }

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{53.1^ \circ } here must be measured clockwise from the direction of west.