Derive the following equations of motion graphically-

1) v = u + at

2) s = ut + 1/2at^{2}

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

@roopasrinath: You can refer to the answers posted by your friends.

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**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**

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**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

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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)

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

Let us suppose that *v *_{0 }is the initial velocity of the particle, at time t = 0, and *v *be the velocity of the particle at t = *t. *

The acceleration *a *of the particle can be written as,

This is the 1 ^{st }kinematic equations for uniformly accelerated motion.

Suppose, the particle travels a distance *x *in the time *t *.

The area of blue colored part gives the distance covered *x *.

So,

This is the 2 ^{nd }equation.

Now, for the 3 ^{rd }equation we shall modify the equation:

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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*+*at*^{2}

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