Equations of Motion

Cricket fan? Hockey fan? Soccer fan? What is the first thing that is taught when you first start training for these or any other sports? It is understanding the correct motion, speed acceleration or the Equations of Motion. Once you master the Equations of Motion you will be able to predict and understand every motion in the world.

Equations of Motion For Uniform Acceleration

As we have already discussed earlier, motion is the state of change in the position of an object over time. It is described in terms of displacement, distance, velocity, acceleration, time and speed. Jogging, driving a car, and even simply taking a walk are all everyday examples of motion. The relations between these quantities are known as the equations of motion.

Equations of Motions

In the case of uniform acceleration, there are three equations of motion which are also known as the laws of constant acceleration. Hence, these equations are used to derive the components like displacement(s), velocity (initial and final), time(t) and acceleration(a). Therefore they can only be applied when acceleration is constant and motion is a straight line. The three equations are,

  • v = u + at
  • v² = u² + 2as
  • s = ut + ½at²

where, s = displacement; u = initial velocity; v = final velocity; a = acceleration; t = time of motion. These equations are referred as SUVAT equations where SUVAT stands for displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (T)

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 Equations of Motion

Derivation of the Equations of Motion

Derivation of First Equation of Motion

  • Simple Algebraic Method:

We know that the rate of change of velocity is the definition of body acceleration.

Let us assume a body that has a mass “m” and initial velocity “u”. Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”. Now we know that:

Acceleration = (Final Velocity-Initial Velocity) / Time Taken

a = v-u /t or at = v-u

v = u + at

  • Graphical Method

Equations of Motion

OD = u; OC = v and OE = DA = t.

Initial velocity = u

Uniform acceleration= a

Final velocity= v

First Equation of Motion:

Let, OE = time (t)

From the graph:

BE = AB + AE

V = DC + OD (QAB = DC & AE = OD)

V = DC + v [QOD = u]

V = DC + V … (1)

Now,

a = (v – u)/ t

a = (OC – OD)/ t = DC/ t

at = DC … (2)

By substituting DC from (2) in (1):

We get:

V = at + u

V = u + at

  • Calculus Method

The rate of change of velocity is acceleration,

a = dv / dt

adt = dv

\int_{0}^{t} adt =\int_{u}^{vdv

at = v-u

v=u +at

Here,

a= acceleration
v=velocity
t=time

u = initial velocity

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Derivation of Second Equation of Motion

  • Simple Algebraic Method:

Let the distance be “s”.

Distance = Average velocity × Time. Also, Average velocity (u+v)/2

Distance (s) = (u+v)/2 × t

Also, from v = u + at

s = (u+u+at)/2 × t = (2u+at)/2 × t

s = (2ut+at²)/2 = 2ut/2 + at²/2

or s = ut +½ at²

  • Graphical Method

equations of motion

OD = u, OC = v and OE = DA = t.

Initial velocity = u

Uniform acceleration= a

Final velocity= v

Distance covered in the given time “t” is the area of the trapezium ABDOE.

Let in the given time (t), the distance covered = s

The area of trapezium, ABDOE.

Distance (s) = Area of ΔABD + Area of ADOE.

s = ½ x AB x AD + (OD x OE)

s = ½ x DC x AD + (u x t) [∵ AB = DC]

s = ½ x at x t + ut [∵ DC = at]

s = ½ x at x t + ut

s = ut + ½ at².

  • Calculus Method

The change rate of displacement is velocity.

It can be equated as:

v = ds/dt

ds = vdt

let’s integrate both the equations together,

ds = (u +at) dt

ds = (u +at) dt=(udt + atdt)

after simplifying it further, we get:

s = ut + ½ at².

Here,

a= acceleration
v=velocity
t=time

u = initial velocity

Derivation of Third Equation of Motion

  • Simple Algebraic Method

We have, v = u + at. Hence, we can write t = (v-u)/a

Also, we know that, Distance = average velocity × Time

Therefore, for constant acceleration we can write: Average velocity = (final velocity + initial velocty)/2 = (v+u)/2

Hence, Distance (s) = [(v+u)/2]  × [(v-u)/a]

or  s = (v² – u²)/2a

or 2as = v² – u²

or v² = u² + 2as

  • Graphical Method

equations of motion

OD = u, OC = v and OE = DA = t

Initial velocity = u

Uniform acceleration= a

Final velocity= v

Distance covered in the given time “t” is the area of the trapezium ABDOE.

Let in the given time (t), the distance covered = s

∴ Area of trapezium ABDOE = ½ x (Sum of Parallel Slides + Distance between Parallel Slides)

Distance (s) = ½ (DO + BE) x OE = ½ (u + v) x t … (3)

Now from equation (2): a=v−ut,

∴ t=v−ua … 4

Now, substitute equation (4) in equation (3) we get:

s= ½ (u+v)×(v−ua),

s = ½a (v + u) (v – u)

2as = (v + u) (v – u)

2as = v² – u²

v² = u² + 2as

  • Calculus Method

a = dv/dt

The second equation that the rate of change in displacement is velocity:

v = ds/dt

by multiplying both the above-written equation,

a ds/dt = v dv/dt

\int_{0}^{s} ads = \int_{u}^{v}vdv

v² = u² + 2as

Here,

a= acceleration
v=velocity
t=time

u = initial velocity

Learn more:

  1. Newtons Law of Motion
  2. Uniform Circular Motion

FAQs on Equations of Motion

Example 1: A body starts from rest accelerate to a velocity of 20 m/s in a time of 10 s. Determine the acceleration of the boy.

Solution: Here, Final velocity v = 20 m/s and initial velocity u = 0 m/s  (the body was at rest yo!). Therefore, Time taken, t = 10 s. Hence, using the equation v = u +at.
a = (v-u )/t
= (20 – 0 ) /10
= 2 m/s2
Hence the acceleration of the body is 2 m/s2.

Example 2: A bus starts from rest and moves with constant acceleration 8ms2. At. the same time, a car travelling with a constant velocity of 16 m/s overtakes and passes the bus. After how much time and at what distance, the bus overtake the car?
A) t =4s, s = 64m     B) t = 5s, s = 72m      C) t = 8s, s = 58m      D) None of These

Solution: A) Let the position of the bus be PB and the position of the car be PC. From s = ut +½ at², we have
Since the initial velocity of the bus, u = 0, hence we have PB = ½ (8)t²
And PC = velocity × Time =  16×t. For the bus to overtake the car, we must have: PB = PC
Hence, ½ (8)t²  = 16×t. Therefore, t = 4s.
Using the value of t = 4s in PB = ½ (8)t ², we have the position of the bus at the time of overtaking is = 64m.

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