Derive the following equations of motion graphically-

1) v = u + at

2) s = ut + 1/2at²

Derivations of Equations of Motion (Graphically)

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

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)

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

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)

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

 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

Suppose 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. s = area of the trapezium OMNQO

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

s = (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

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

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