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## MOTION IN A PLANE WITH CONSTANT ACCELERATION

Suppose that an object is moving in x-y plane and its acceleration a is constant. Over an interval of time, the average acceleration will equal this constant value. Now, let the velocity of the object be v0 at time t = 0 and v at time t. Then, by definition

`a=(v-v_0)/(t-0)=(v-v_0)/t`

Or, `v=v_0 + at`

In terms of components:

`v_x=v_(0_x) + a_xt`

`v_y=v_(0_y) + a_yt`

Let us now find how the position r changes with time. We follow the method used in the one dimensional case. Let ro and r be the position vectors of the particle at time 0 and t and let the velocities at these instants be vo and v. Then, over this time interval t, the average velocity is (vo + v)/2. The displacement is the average velocity multiplied by the time interval :

`r-r_0=((v+v_0)/2)t=(((v_0+at)+v_0)/2)t=v_0t+1/2at^2`

Or, `r=r_0 +v_0t+1/2at^2`

It can be easily verified that the derivative of the above equation satisfies the condition that at t=0, `r=r_0`. The above equation can be written in component form as

`x=x_0 + v_(0_x)t + 1/2 a_xt^2`

`y=y_0 + v_(0_y)t + 1/2 a_yt^2`

One immediate interpretation of the above equation is that the motions in x- and y-directions can be treated independently of each other. That is, motion in a plane (two-dimensions) can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions.