Equations of Motion for Uniform Accelerated Motion Physics Notes

Equations of Motion for Uniform Accelerated Motion:
In uniform acceleration, magnitude and direction of an object always remains constant. This type of motion is called uniform accelerated motion.

For uniform accelerated motion, the graph between velocity and time is a straight line and its slope is equal to the acceleration.

The area under the velocity-time graph expresses the distance or displacement.

For the study of uniform accelerated motion, the equation which shows the relation among velocity, time, distance and acceleration is called the equation of motion.

NCERT Solutions Guru Equations of Motion for Uniform Accelerated Motion Physics Notes

There are three equations of motion,
1. Graphical method: Consider an object moving along a straight line with uniform acceleration a. Let u be the initial velocity of the object at time t = 0 and u be the final velocity of the object at time t. Let s be the distance travelled by the object in time t.
Equations of Motion for Uniform Accelerated Motion Physics Notes 1
Equation of Motion by the Graphical Method

The Velocity-time graph of this motion is a straight line PQ, as shown in the figure.
where, OP = u = RS
OW = SQ = v
and OS = PR = t
1. For the first equation of motion: We know that the slope of velocity-time graph of uniformly accelerated motion represents the acceleration of the object.
i.e., Acceleration = slope of the velocity-time graph PQ
or a = \(\frac{Q R}{P R}=\frac{Q R}{O S}\)
a = \(\frac{S Q-S R}{O S}=\frac{v-u}{t}\)
or v – u = at
or v = u + at …….(i)
This is the first equation of uniform accelerated motion.

NCERT Solutions Guru Equations of Motion for Uniform Accelerated Motion Physics Notes

2. Second equation of motion: We know that the area under the velocity-time graph for a given time interval represents the distance covered by the uniformly accelerated object in that interval of time.

Distance (displacement) travelled by the object in time t is :
S = area of trapezium OSQP
= area of rectangle OSRP + Area of triangle PRQ
or S = OS × OP+ \(\frac{1}{2}\) × PR × RQ
(Area of rectangle = Length × Breadth)
(Area of triangle = \(\frac{1}{2}\) × Base × Height)
= t × u + \(\frac{1}{2}\) × t × (v – u)

(From the first equation of motion v – u = at)
= ut + \(\frac{1}{2}\) × t × at
Thus, S = ut + \(\frac{1}{2}\) at2 ……..(iii)
This is the second equation of uniform accelerated motion.

3. Third equation of motion : Distance travelled by the object in time interval t is
s = area of trapezium OSQP
= \(\frac{1}{2}\) (OP + SQ) × OS
OP = SR
= \(\frac{1}{2}\)(SR + SQ) × OS …(iii)

Acceleration, a = slope of the velocity-time graph PQ
Equations of Motion for Uniform Accelerated Motion Physics Notes 2
This is the third equation of uniform accelerated motion.

NCERT Solutions Guru Equations of Motion for Uniform Accelerated Motion Physics Notes

2. Calculus method:
1. Velocity-time relation: These equations can also be derived from the calculus method. From the definition of acceleration;
a = \(\) or dv=adt

Integrating it within the condition of motion (i.e.) when the time changes from 0 to t, velocity changes from u to u, we get
Equations of Motion for Uniform Accelerated Motion Physics Notes 3
This is the first equation of motion.

2. Distance time relation : The instantaneous velocity of an object in uniformly accelerated motion is given by
v = \(\frac{d x}{d t}\) or dx = vdt
dt
Now v = u + at
∴ dx =(u + at)dt …(vii)
Let the displacement of the object from the origin of position-axis is x0 at t = 0 and x at t = t, integrating both the sides of the equation (vii) within proper limits, we have,
Equations of Motion for Uniform Accelerated Motion Physics Notes 4
or x – xo = u(t – 0) + a(\(\frac{t^{2}}{2}\) – 0]
or x—x0 = ut + \(\frac{1}{2}\)at …(viii)
If x – x0 = S = distance covered by the object in time t, then from eq. (viii)
s = ut + \(\frac{1}{2}\)at2 …(ix)

This is the second equation of uniform accelerated motion.

NCERT Solutions Guru Equations of Motion for Uniform Accelerated Motion Physics Notes

3. Velocity-displacement relation
The instantaneous acceleration is given by
Equations of Motion for Uniform Accelerated Motion Physics Notes 5
Let u and v be the velocity of the object at positions given by displacements x0 and x.

Integrating the above equation (x) with the condition of motion, we get
Equations of Motion for Uniform Accelerated Motion Physics Notes 6
If x – x0 = s = the distance covered by the object in time t, then from eq. (xi),
\(\frac{1}{2}\)(v2 – u2) = as
or v2 – u2 = 2as
or v2 = u2 + 2as …..(xii)
This is the third equation of uniform acceleration motion

Distance travelled by an object in nth second: We know that the distance travelled by an object iii time t is
s = ut + \(\frac{1}{2}\)at2 ……(i)
in n sec; s = un + \(\frac{1}{2}\)an2 …….(ii)

Similarly distance travelled in (n – 1) sec
Equations of Motion for Uniform Accelerated Motion Physics Notes 7
Kinetic Equation of Motion under Gravity:
When an object falls freely under the effect of gravity, it is accelerated towards the centre of the earth with an acceleration of 9.8 m/sec2 (or 980 cm s-2) called the acceleration due to gravity (g). The motion of an object falling freely under gravity is, thus, a case of motion with uniform acceleration.

The kinetic equation of motion under gravity can be obtained by replacing ‘a’ by ‘g in the of motion [obtained earlier. Accordingly, the kinematic: equations of motion under gravity are as below:

  • v = u ± gt
  • S = ut ± \(\frac{1}{2}\)gt2
  • v2 = u2 ± 2gs
  • Sn = u ± \(\frac{1}{2}\) g(2n – 1)

For upward motion: a = g take
For downward motions: a = +g

In the above expression, resistance due to air has been neglected. Further, when an object, falls freely, its initial velocity u is zero and the value of ‘g is positive. On the other hand, when an object is thrown up against gravity, it will rise till its final velocity u becomes zero. In this case, the value of ‘g is negative.

Physics Notes