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Graphical analysis of motion with constant acceleration in one dimension is based on the equation defining position and displacement, which is a quadratic function. Nature of motion under constant acceleration and hence its graph is characterized by the relative orientation of initial velocity and acceleration. The relative orientation of these parameters controls the nature of motion under constant acceleration. If initial velocity and acceleration are in the same direction, then particle is accelerated such that speed of the particle keeps increasing with time. However, if velocity and acceleration are directed opposite to each other, then particle comes to rest momentarily. As a result, the motion is divided in two segments - one with deceleration in which particle moves with decreasing speed and second with acceleration in which particle moves with increasing speed.

We study various scenarios of motion of constant acceleration by analyzing equation of position, which is quadratic expression in time. The position of particle is given by :

x = x 0 + u t + 1 2 a t 2 x = 1 2 a t 2 + u t + x 0

Clearly, the equation for position of the particle is a quadratic expression in time, “t”. A diagram showing a generalized depiction of motion represented by quadratic expression is shown here :

Motion under constant acceleration

Motion under constant acceleration

In case initial position of the particle coincides with origin of reference, then initial position x 0 = 0 and in that case x denotes position as well as displacement :

x = u t + 1 2 a t 2

Nature of graphs

The nature of quadratic polynomial is determined by two controlling factors (i) nature of coefficient of squared term, t 2 and (ii) nature of discriminant of the quadratic equation, which is formed by equating quadratic expression to zero.

Nature of coefficient of squared term

The coefficient of squared term is “a/2”. Thus, its nature is completely described by the nature of “a” i.e. acceleration. If “a” is positive, then graph is a parabola opening up. On the other hand, if “a” is negative, then graph is a parabola opening down. These two possibilities are shown in the picture.

Motion under constant acceleration

Motion under constant acceleration

It may be interesting to know that sign of acceleration is actually a matter of choice. A positive acceleration, for example, is negative acceleration if we reverse the reference direction. Does it mean that mere selection of reference direction will change the nature of motion? As expected, it is not so. The parabola comprises two symmetric sections about a line passing through the minimum or maximum point (C as shown in the figure). One section represents a motion in which speed of the particle is decreasing as velocity and acceleration are in opposite directions and second section represents a motion in which speed of the particle is increasing as velocity and acceleration are in the same direction. In other words, one section represents deceleration, whereas other section represents acceleration. The two sections are simply exchanged in two graphs. As such, a particular motion of acceleration is described by different sections of two graphs when we change the sign of acceleration. That is all. The nature of motion remains same. Only the section describing motion is exchanged.

Questions & Answers

how did you get the value of 2000N.What calculations are needed to arrive at it
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James Reply
can anyone tell who founded equations of motion !?
Ztechy Reply
n=a+b/T² find the linear express
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Sultan Reply
Moment of inertia of a bar in terms of perpendicular axis theorem
Sultan Reply
How should i know when to add/subtract the velocities and when to use the Pythagoras theorem?
Yara Reply
Centre of mass of two uniform rods of same length but made of different materials and kept at L-shape meeting point is origin of coordinate
Rama Reply
A balloon is released from the ground which rises vertically up with acceleration 1.4m/sec^2.a ball is released from the balloon 20 second after the balloon has left the ground. The maximum height reached by the ball from the ground is
Lucky Reply
work done by frictional force formula
Sudeer Reply
Misthu Reply
Why are we takingspherical surface area in case of solid sphere
Saswat Reply
In all situatuons, what can I generalize?
Cart Reply
the body travels the distance of d=( 14+- 0.2)m in t=( 4.0 +- 0.3) s calculate it's velocity with error limit find Percentage error
Clinton Reply
Explain it ?Fy=?sN?mg=0?N=mg?s
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