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Motion in a Plane with Constant Acceleration

The average velocity of the object in the time interval between t and t’ is given by
=                  ….(4.1)
 
As said earlier, while studying uniformly accelerated motion in a straight line, is not the actual velocity of the object during the time interval t’ – t. It is neither the velocity of the object at time t nor at t’ or at any other instant between t and t’.
If =are velocities at times t and t’ respectively, then constant acceleration of the motion of the object is given by
=                    …..(4.2)
 
This value of acceleration is same for all values of t and t’. In each unit of time, the velocity of the object will always change by an amount equal to the magnitude of acceleration given by equation (4.1) and always in the same direction.

Suppose that at t = 0, the velocity of the object is . If at time t = 0, ux and uy are component of the velocity of the object along X-axis and Y-axis respectively, then
= ux+ uy             ….. (4.3)
 
As the object is moving with constant acceleration , the velocities of object at times t and t’
=                   …..(4.4)
and =            …..(4.5)
 
Subtracting equation(4.4)from the equation (4.5), we have
=
= +(t’ – t)           …(4.6)
or
the above equation can also be obtained from equations (4.2) by expanding it.
If ax and ay are magnitudes of the components of acceleration along X-axis and Y-axis respectively then
= ax + ay             …..(4.7)
such that
 
 
From equation (4.3) and (4.7), substituting for and in the equation (4.4), we have
= (ux+ uy) + (ax+ ay) t
 
If vx and vy are magnitudes of the component of velocity of time t, then
vx+ vy = (ux+ uy) + (ax+ ay) t
or vx+ vy = (ux+ axt) + (ux+ ayt)
 
The above equation will hold only, if the coefficients of and on the two sides of the above equation are separately equal i.e. if
vx = ux + axt                 … (4.8)
and vy = uy + ayt           … (4.9)
 
In order to visualize how the object moves in a plane when its motion has constant acceleration, let us obtain for the x and y coordinates of the position of the object at any time t. If is the average velocity of the object in time interval between t and t’, then according to equation (4.1), the position of the object at time t’ is given by
= (t’ – t)               ….(4.10)
Also, =
 
Substituting the value of in equation (4.10), we have
= (t’ – t)
=       …..(4.11)
 
The equation (4.11) connects the position of the object at any time t’ to its position at any time t, when moving with constant acceleration . The position at any time t can be related to position at t = 0 by changing t’ to t, t to O, to and to in equation (4.11) and likewise, we have
=
 
In terms of rectangular components, the above equation may be written as
x+ y = (x0+ y0) + (ux+ uy)t + ) t2
or x+ y = (x0+ uxt+ + (y0+ uyt +
 
The above equation will hold only, if the coefficients of and on the two sides of the above equation are separately equal i.e., if
x = x0 =uxt+              …(4.13)
and y = y0 =uyt+        …(4.14)
 
The above two equations can be used to see how the x and y coordinates position of the object change with passage of time. If we eliminate t from equations (4.13) and (4.14), we shall obtain the equation of path followed by the object moving in a plane with constant acceleration. However, the path can be traced * for given values of x0, y0, ux, uy, ax and ay.

Let us consider a particular case of throwing an object in a vertical XY-plane with some velocity making an angle θ with horizontal from point say P(x0 , y0) in the plane. In such a case, ax = 0 and ay = -9.8 ms-2. For certain values of ux and uy, the values of x and y at time t = 1s, 2s, 3s,… can be found. If we plot a graph between x and y, the graph will be a parabola as shown in the figure.




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