Question

A man is sitting in a car moving at 6 m/s towards North. As seen by him, a cyclist is moving towards East at 4.5 m/s Then, the speed of the cyclist as seen by an observer on the ground is:
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Answer 1

Answer:

Explanation:

Let say Cyclist Speed was x m/s  North  y° East

Then cyclist speed in North Direction = x Cosy°

Cyclist speed in East Direction = xSiny°

As relative speed with 6m/s toward north is  East at 4.5 m/s

=> x Cosy° = 6   ( north Directions balancing each other)

& x siny° = 4.5  

Squaring both and adding

x²Cos²y° + x²Sin²y° = 6² + 4.5²

=> x² = 56.25

=> x = 7.5 m/s

x siny° = 4.5  

=> 7.5 Siny° = 4.5  

=> 5Siny° = 3

=>Siny° = 3/5

=> y° = 36.87°

7.5 m/s  North 36.87° East

Answer 2

A man is sitting in a car moving at 6m/s towards North.

in vector form, velocity of man, $\vec{v_m}=6\hat{j}$

as seen by man, A cyclist is moving towards East at speed 4.5 m/s.

in vector form, velocity of cyclist with respect to man, $\vec{v_{c,m}}=4.5\hat{i}$

we know, $\vec{v_{c,m}}=\vec{v_c}-\vec{v_m}$

or, $4.5\hat{i}=\vec{v_c}-6\hat{j}$

or, $4.5\hat{i}+6\hat{j}=\vec{v_c}$

hence, velocity of cyclist with respect to observer on the ground is $4.5\hat{i}+6\hat{j}$

and magnitude of speed of cyclist is √{4.5² + 6²} = 7.5 m/s