Question

Derive first equation of motion?​

Answer 1

Motion and Rest

If the position of an object changes with respect to a reference point then it is said to be in motion wrt.that reference while if it does not changes then it is at rest wrt.that reference point. For the better understanding or to deal with the different situations of rest and motion we derive some standard equation relating terms distance,displacement,speed,velocity and acceleration of the body by the equation called as equations of motion.

Three Equations of Motion

In case of motion with uniform or constant acceleration (one with equal change in velocity in equal interval of time) we derive three standard equations of motion which are also known as the laws of constant acceleration. These equations contain quantities displacement(s), velocity (initial and final), time(t) and acceleration(a) that governs the motion of a particle. These equations can only be applied when acceleration of a body 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.

Derivation of Equation of Motions

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Now let's start the derivation with the first equation of motion i.e. v=u+at where u is the initial velocity, v is the final velocity and a is the constant acceleration.

Assuming that a body started with initial velocity “u” and after time t it acquires final velocity v due to uniform acceleration a.

We know acceleration is defined as the rate of change of velocity, also which is given by slope of the velocity time graph.

Thus both from definition as well as graph Acceleration = Change in velocity/Time Taken i.e. a = v-u /t or at = v-u

Therefore, we have: v = u + at

Now to derive the second equation again suppose a body is moving with initial velocity u after time t its velocity becomes v. The displacement covered by the during this interval of time is S and the acceleration of the body is represented by a.

Explanation: We know area under velocity time graph gives total displacement of the body thus area under velocity time graph is area of trapezium OABC.

Also area of trapezium = ½(sum of parallel sides)height

Sum of parallel sides=OA+BC=u+v and here,height=time interval t

Thus,area of trapezium = ½(u+v)t

Substituting v=u+at from first equation of motion we get,

Displacement =S =area of trapezium = ½(u+u+at)t

S = ½(2u+at)t=ut+½at2

Which is called the second equation of motion and is the relation between displacement S,initial velocity u,time interval t and acceleration a of the particle.

Now in order to derive the third equation again use

Displacement =S =area of trapezium = ½(u+v)t

From first equation v=u+at we get v-u=at ⇒v-u/a=t

Substituting the value of t in S = ½(u+v)t

We get S=½(u+v)(v-u)/a=(v2-u2)/2a

⇒2as=v2-u2

⇒v2 =u2+2as

Which is the third equation of motion and is the relation between final velocity v,initial velocity u,constant acceleration a and displacement S of the particle.

We can now also calculate the displacement of particles during the nth second, using these equations of motion derived above. In order to do so we will calculate the displacement covered in n seconds and subtract the displacement covered in n-1 seconds and to get the displacement in nth second

snth =Sn-sn-1=un-u(n-1) + 1/2an2-1/2a(n-1)2

simplifying gives us final equation for displacement in the nth second is s = u + a(2n-1)/2

This equation is often regarded as modified form of second equation of motion

Answer 2

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Let's start the derivation with the first equation of motion,

Let us consider an object moving in a straight line with a constant acceleration. For such a situation the first equation of motion gives the final velocity (v) after time t given that the object is having an initial velocity of u and constant acceleration of a. Now for deriving it, let us consider the following velocity-time graph:

The initial velocity of the body, u=OA (1)

The final velocity of the body, v=BC (2)

From the graph BC = BD + DC

Therefore, v = BD+DC (3)

Again DC=OA

So, v=BD+OA

Now from equation (1), OA=u

So, v=BD+u (4)

We should find the value of BD now

Now, from the velocity-time graph, if we calculate the change in velocity to change in time, then we can find the acceleration, a. This is nothing but the slope of the graph.

Thus,

Acceleration, a = Slope of line AB

a=

$\frac{bd}{ad}$

But AC=OC=t

Hence we get,

$\frac{bd}{t}$

bd = at

Now putting this in equation (4) we get

v = u + at

hope it helps u.