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Points B, D : The curve intersects t-axis. The particle is at the origin of the reference, O, chosen for the motion. The time corresponding to these points (B and D) are real roots of the quadratic equation, which is obtained by equating quadratic expression to zero.

Point C : The slope of tangent to the curve at this point is zero. It means that the speed of the particle has reduced to zero. The particle at this point is at rest. The slopes of the tangent to curve about this point changes sign. It means that velocity is oppositely directed about this point. The point B, therefore, is a point, where reversal of direction of motion occurs. Note that particle can reverse its direction only once during its motion under constant acceleration.

Point E : The particle reaches start point E (i.e. A) again in its motion after reversal of motion at C.

Point F : This is the end point of motion.

The graph of a motion under constant acceleration is bounded by set of parameters defining a motion. In a particular case, we may be considering only a segment of the curve starting from point A and ending at point F – not necessarily covering the whole of graph as shown in the figure. The point F may lie anywhere on the graph. Further, nature of curve will be determined by values and signs of various parameters like initial position, initial velocity and acceleration. Here, we have divided our study in two categories based on the sign of acceleration : (i) acceleration is positive and (ii) acceleration is negative.

Acceleration is positive (in the reference direction)

The graph of quadratic equation is a parabola opening upwards as coefficient of squared term t 2 is positive i.e. a>0. The minimum value of expression i.e. x is :

x min = D 4 A = u 2 2 a x 0 4 X a 2 = u 2 2 a x 0 2 a

The various possibilities of initial positions of the particle with time are discussed here :

Case 1: Initial position and origin of reference are different. Initial position is positive. Initial velocity is negative i.e. it is directed in negative reference direction. The velocity and acceleration are oppositely directed.

Motion under constant acceleration

Motion under constant acceleration

Initially particle is decelerated as the speed of the particle keeps on decreasing till it becomes zero at point C. This is indicated by the diminishing slope (magnitude) of the tangents to the curve. Subsequently, particle is accelerated so long force causing acceleration is applied on the particle.

The segment DF with origin at D is typical graph of free fall of particle under gravity, considering DF as the height of fall and downward direction as positive direction. Only differing aspect is that particle has initial velocity. Nevertheless, the nature of curve of free fall is similar. Note that speed of the particle keeps increasing till it hits the ground.

Problem : Given x 0 = 10 m ; u = - 15 m / s ; a = 10 m / s 2

Find the position when particle reverses its motion.

Solution : Acceleration is negative. The position-time graph is a parabola opening up. Therefore, the particle changes its direction of motion at its minimum position, which is given as :

x min = - u 2 2 a x 0 2 a

x min = - 15 2 2 X 10 X 10 2 X 10 = 25 20 = 1.25 m

Alternatively,

At the point of reversal, velocity is zero. Using v=u+at, :

0 = - 15 + 10 t t = 3 2 = 1.5 s

Position of particle in 1.5 s is :

Motion diagram

Motion diagram

x = x 0 + u t + 1 2 a t 2 x = 10 + 15 X 1 .5 + 1 2 X 10 X 2.25 = 10 22.5 + 5 X 2.25 = 12.5 + 11.25 = 1.25 m

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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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