# Relative motion involving angle

Suppose I am going in a direction with a velocity $v_1$ and my friend is going in a direction which makes an angle of $A$ with my direction with a velocity of $v_2$.

Then what will be my relative motion with respect to my friend or his relative motion with respect to me?

Every book I have read gave example where two things are going parallel or opposite of me. But I've never found any example with an angle.

• @AsifIqubal...You draw the vectors of yours and friend's first .Then you also draw the opposite vector of your friend.Taking into consideration the vector of yours and friend with angle π-A , put in the resultant equation.This is the relative velocity... – Nehal Samee Feb 21 '18 at 13:08
• Are you traveling on a flat surface, or in 3 dimensional space? – David White Apr 26 '19 at 18:58

Let's say your velocity is $$\vec{v}_1$$ is $$\vec{v}_1= v_{1x}\hat{i} + v_{1y}\hat{j}$$ and your friends velocity $$\vec{v}_2$$ is $$\vec{v}_2= v_{2x}\hat{i}+v_{2y}\hat{j}$$ Then your velocity relative to your friend will be $$\vec{v}_r = (v_{1x}-v_{2x})\hat{i} + (v_{1y}-v_{2y})\hat{j}$$ The components can be found by considering an appropriate coordinate system. In this case that will be a coordinate system with the $$x$$-axis aligned with your friends velocity. In that case $$v_{2y} =0$$. While the components of your velocity will be $$v_{1x}=v_{1}\cos{A}$$ and $$v_{1y}=v_{1}\sin{A}$$.

Hope this helps.

You could always break the question into two perpendicular directions

Your velocity will be subtracted from horizontal velocity of your friend's. The vertical velocity will remain same. The resultant of the vertical and new horizontal velocity will give you the final answer of how you see your friend.

Velocity of friend in horizontal direction will be $(v_2 \cos A)$
Velocity of friend in vertical direction will be $(v_2 \sin A)$

Horizontal Direction:
$(v_2 \cos A -v_1)$ is the velocity of your friend in the horizontal direction relative to you.

Vertical Direction:
$(v_2 \sin A)$ is the velocity of your friend in the vertical direction relative to you.
Nothing changes as your entire velocity is in the horizontal direction.

Combining the two directions: $\sqrt{(v_2 \cos A -v_1)^2 + (v_2 \sin A)^2}$ is the velocity of your friend relative to you.

Method 2
The resultant vector of your friends velocity vector with the negative of your velocity vector
i.e. (Friends velocity vector) - (your velocity vector)

• Can you plaese elaborate and express yourself in equations?I didn't understand what you said.. – Theoretical Feb 21 '18 at 12:35
• Why is my entire velocity in the horizontal direction? – Theoretical Feb 22 '18 at 8:37
• @AsifIqubal Say your going towards the right, your entire velocity is horizontal. You are not going up. If your velocity is towards North-East, your velocity has a vertical component and a horizontal component. In the question I assume you are going purely horizontal while your friend is at an angle. – SmarthBansal Feb 22 '18 at 10:48
• Is it legal? I mean that will it work if you guess my velocity to be purely horizontal. I mean I can go towards Northeast as you said. Then how would you describe that? – Theoretical Feb 22 '18 at 11:18
• I am free to choose the perpendicular directions. I can now view the problem from southeast and if you are going north east i can still see you going right. If you don't like the shift of axis then you would need to break your velocity too in the horizontal and vertical directions and then subtract it from your friend's horizontal,vertical direction, then combine up the result. – SmarthBansal Feb 22 '18 at 11:27

Using the notation that $\vec v_{\rm ab}$ is the velocity of $a$ relative to $b$ you need the velocity of $1$ relative to $2$ which is $$\vec v_{12} = \vec v_{1\rm g} + \vec v_{\rm g 2} = \vec v_{1\rm g} - \vec v_{\rm 2 g}$$

which diagrammatically is as follows with $g$ as the ground.

It might help remember how to set out the vector addition by looking at the sequence of the subscript symbols

$1\,2 = 1\,\mathbf{g} + \mathbf{g}\, 2$

• I didn't understand how your answer helps me in my question? – Theoretical Feb 21 '18 at 14:04
• You know two sides of a triangle $\vec v_{1\rm g}$ and $\vec v_{2\rm g}$ (the velocities of you and your friend relative to the ground) and the angle between them $A$ so you can use trigonometry to find the other velocity $\vec v_{12}$ (velocity of you relative to your friend). – Farcher Feb 21 '18 at 14:32
• How can I do that? – Theoretical Feb 22 '18 at 8:42