# Derive the following equations of motion graphically-1) v = u + at2) s = ut + 1/2at2

@vknov21, @martyr : Good work! Keep it up!

• 4

## First Equation of Motion Graphical Derivation of First Equation

Consider an object moving with a uniform velocity u in a straight line. Let it be given a uniform acceleration a at time t = 0 when its initial velocity is u. As a result of the acceleration, its velocity increases to v (final velocity) in time t and S is the distance covered by the object in time t.

The figure shows the velocity-time graph of the motion of the object.

Slope of the v - t graph gives the acceleration of the moving object.

Thus, acceleration = slope = AB = ##### v - u = at

v = u + at I equation of motion

• 8

ii) Graphical Derivation of Second Equation

Distance travelled S = area of the trapezium ABDO

= area of rectangle ACDO + area of DABC     (v = u + at I eqn of motion; v - u = at) ##### • 6

First equation of motion:

Consider a body having initial velocity u,If it is subjected to an acceleration 'a' so that after time 't' its velocity becomes v.

By the definition of acceleration

acceleration=Change in velocity/time taken

a=(v-u)/t

v=u+at

Second equation of motion:

Suppose the distance travelled by the above body in time t=s

we have average velocity=(Initial velocity+Final velocity)/2 • 11

Suppose a body has an initial velocity U and is accelerating at the rate 'a'. The distance travelled in the time t = (n-1) second is,

Using,

S = Ut + 1/2 at 2

=> [S] n-1 = U(n-1) + 1/2 a(n-1) 2

=> [S] n-1 = Un – U + 1/2 a(n 2 + 1 – 2n)

=> [S] n-1 = Un – U + 1/2 an 2 + 1/2 a - an

The distance travelled by the body in time t = n second is,

S = Ut + 1/2 at 2

=>[S] n = Un + 1/2 an 2

Now,

The distance travelled in the n th second is given by,

S n = [S] n – [S] n-1

=> S n = Un + 1/2 an 2 - (Un – U + 1/2 an 2 + 1/2 a – an)

=> S n = U + an – 1/2 a

=> S n = U + a(n – 1/2)

=> S n = U + (a/2)(2n - 1)

• -5

Here is the derivation of the equations of motion by graphical method. Follow the graph:

Let us suppose that is the initial velocity of the particle, at time t = 0, and be the velocity of the particle at t = t.

The acceleration of the particle can be written as, This is the 1 st kinematic equations for uniformly accelerated motion.

Suppose, the particle travels a distance in the time . The area of blue colored part gives the distance covered .

So, This is the 2 nd equation.

Now, for the 3 rd equation we shall modify the equation: • -1

We can relate the velocity of a uniformly accelerating body with the distance covered by it in a certain time. This is called a one-dimensional position−time equation of a body moving with uniform acceleration. This equation is used to obtain the distance travelled by the body within a given interval of time.

Position−time relation from velocity−time graph

S uppose a body is moving in a straight path with initial velocity u and constant acceleration a, such that it covers a distance s in time t. The given figure shows the velocity−time graph that represents the motion of the given body.

Length MO represents the initial velocity,

i.e. MO = u

{Note that in many cases where the body starts from rest, initial velocity u = 0. However, in this case, u ≠ 0}

The straight line MN represents the velocity-time curve.

Let distance travelled by the body in time t be s.

Now, the area enclosed by the line MN with the time axis gives the total distance travelled by the object.

i.e. = area of the trapezium OMNQO

= (area of the rectangle OMQP) + (area of the triangle MNP)

= (OM × MP) + ( × MP × PN )

Since NQ represents the change in velocity v after time t, using the first equation of motion i.e. v = u+at, we obtain

NQ = u+at

PQ = u and NQ = NP + PQ, ( from the graph)

Hence, NP = at,

And, MP = t

∴ s = (u × t) + ( × at × t)

 s = ut + at2
• 6

Thnxx, everyone.....helped a lot...

• -3

Question:derive the equation : S=ut+1/2at^2 from velocity time graph for a uniformly accelerated motion.

• -5

Question: how many points two distinct lines can intersect?

• 1

What are concurrent angles?

• -2
What are you looking for?