Physics Motion in Plane

  $\left|\widehat{A}+\widehat{B}\right|=\sqrt[]{A^2+B^2+2ABcos\theta }$ $\stackrel{}{}\stackrel{}{}\text{exception 25:}$

$=\sqrt[]{1+1+2cos\theta }$

$=\sqrt[]{2\left(1+cos\theta \right)}$

= $\sqrt[]{2*2co{s}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ $\text{exception 25:}$

$$ $\left|\widehat{A}+\widehat{B}\right|=2cos\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  -(1)

$\left|\widehat{A}-\widehat{B}\right|=\sqrt[]{A^2+B^2-2ABcos\theta }$

$=\sqrt[]{1+1-2cos\theta }$

$\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{$$}\right.$ $=2sin\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  -(2)

(2) $÷$ $$ (1)

$\frac{\left|\widehat{A}-\widehat{B}\right|}{\left|\widehat{A}+\widehat{B}\right|}=\frac{2sin\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2cos\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\Rightarrow \left|\widehat{A}-\widehat{B}\right|=\left|\widehat{A}+\widehat{B}\right|tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$\overrightarrow{A}*\overrightarrow{A}=AAsin\theta \widehat{n}$

$=AAsin0°\widehat{n}$

= 0 [since Angle between the vectors are zero degree]

$\overrightarrow{A}*\overrightarrow{A}=0$

In 1D kinematics, you use scalar equations for one direction. In 2D, position, velocity, and acceleration become vectors with x and y components. You apply the same kinematic equations independently to each dimension. Just remember to treat horizontal and vertical motions as separate 1D problems to be solved simultaneously.

In uniform circular motion, we know that the speed is constant. But the velocity vector's direction continuously changes as the object moves in a circle. This continuous change in direction leads to an acceleration. In physics, we call that centripetal acceleration. This is always directed towards the centre of the circle.

When we speak of 2D or 3D motion, the velocity and acceleration vectors need not align or be in the same direction. They can have any angle between 0° and 180° between them. This is because acceleration accounts for changes in both the magnitude and direction of velocity.

Vectors are necessary because motion in a plane (two dimensions) or space (three dimensions) involves physical quantities, including velocity and acceleration. Both of these have both magnitude and direction. That helps us know how objects move in the real-world and in any real space. In one-dimensional motion, we only can know two directions, and show it as signs (+/-), and not more than that.  Motion in a plane requires vectors to accurately represent these directional aspects.

Path length is a scalar quantity. It tells about the total distance covered, even though it depends on the path taken. Displacement is a vector quantity.  It tells that it's a straight-line change in position from the initial to the final point. Now, that is path independent. 

To find the direction in vector subtraction, let's consider (A – B).  We have to use vector addition by rewriting it as A + (–B). Then, this negative vector (–B) will have the same magnitude as B. Only that it will point in the opposite direction. Then we will use the head-to-tail method. Following that, we will place the tail of (–B) at the head of A. The resultant vector from the tail of A to the head of –B will give us both the magnitude and direction of A – B.