Question

derive equations of motion for a particle moving in a plane and show that the motion can be resolved in two independent motions in mutually perpendicular directions​

Answer 1

Answer:

Let the initial velocity of the particle be

$\vec u=u_x\hat i+u_y\hat j$

The acceleration of the particle be

$\vec a=a_x\hat i+a_y\hat j$

After time t if the velocity is $\vec v=v_x\hat i+v_y\hat j$ and the displacement is $\vec s$

Then

Acceleration = rate of change of velocity

$\vec a=\frac{\vec v-\vec u}{t}$

$\implies \vec a \times t=\vec v-\vec u$

$\implies \boxed{\vec v=\vec u+\vec a t}$        ............ (1)

Average velocity

$=\frac{\vec u+\vec v}{2}$

$\vec s=\frac{\vec u+\vec u+\vec a t}{2}\times t$

$\implies \vec s=\frac{2\vec u t+\vec a t^2}{2}$

$\implies \boxed{\vec s=\vec u t+\frac{1}{2}\vec a t^2}$   ............(2)

Taking the dot product of eq (1) with itself

$\vec v.\vec v=(\vec u+\vec a t).(\vec u+\vec a t)$

$\implies \vec v.\vec v=\vec u.\vec u+2\vec u.\vec a t+\vec a.\vec a t^2$

$\implies \vec v.\vec v=\vec u.\vec u+2\vec a.(\vec u t+\frac{1}{2}\vec a t^2)$

$\implies \boxed{\vec v.\vec v=\vec u.\vec u+2\vec a.\vec s}$  ........ (3)

(1), (2) and (3) are required equations of motion

From equation (1)

$\vec v=\vec u+\vec a t$

$v_x\hat i+v_y\hat j=(u_x\hat i+u_y\hat j)+(a_x\hat i+a_y\hat j) t$

$\implies v_x\hat i+v_y\hat j=(u_x\hat i+a_x\hat i t)+(u_y\hat j+a_y\hat j t$

Comparing the x and y component

$v_x=u_x+a_xt$

And

$v_y=u_y+a_yt$

From equation (2)

$\vec s=\vec u t+\frac{1}{2}\vec a t^2$

$s_x\hat i+s_y\hat j=(u_x\hat i+u_y\hat j) t+\frac{1}{2}(a_x\hat i+a_y\hat j) t^2$

$\implies s_x\hat i+s_y\hat j=(u_x t+\frac{1}{2}a_x t^2)\hat i+(u_y t+\frac{1}{2}a_y t^2)\hat j$

Comparing the x and y components

$s_x=u_x t+\frac{1}{2}a_x t^2$

$s_y=u_y t+\frac{1}{2}a_y t^2$

Similarly for equation (3) we can say that the motion can be resolved in two independents mutually perpendicular directions.

Hope this helps.