Equations of Motion - Derivation of Three Equations and Under Gravity

For a Uniformly Accelerated Motion:

Suppose an object is moving in a straight line under constant acceleration. In that case, the relationship between its velocity, time, acceleration, and displacement can be represented by three equations which are termed equations of motion.

Let an object move in a straight line under constant acceleration, cover distance ‘s’ in time t, and let its velocity be v. The three equations can be obtained by the graphical method.

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

But the equations will be accurate only when the acceleration is constant and the motion is along a straight line.

Derivation of the Three Equations

The initial velocity of the body at time t=0 is u. After that, it increases continuously at a uniform rate as acceleration is kept constant (a). The time-velocity graph of the object is an inclined straight line, AB, and after time t, the velocity becomes v. The coordinates of point B on AB are (t,v).

1^st equation (v=u + at):  Slope of time-velocity graph gives acceleration. 

So, a= CB/AC = (v-u)/t

=>V-u=at 

=> v=u+at 

2^nd equation (S= ut + ½ at²): The object covers distance s from time t=0 to time t=t which will be equal to the area enclosed between the time-velocity graph and time axis from t=0 to t.

S= area ABDO

= Area of the rectangle ACDO + area of triangle ABC

            =(DC*AC) + ½ (CB*AC) = ut + ½ (v – ut) *t

          Again, from the 1^st equation of motion, v – u = at

                            So, s=ut + ½ (at)*t

S=ut + ½ at²


 

3^rd equation (v² = u² + 2as) : The distance travelled by a  moving  object in time t is given by ,

            S= Area of trapezium ABCDO

               = sum of parallel sides * the distance between them

               = [(OA + DB) *OD]/2= ½ (u + v) * t

Again, t= (v – u)/a

S=1/2 (u+v) *[ (v – u)/a] = (v^{2 _} u²)/2a
V² = u² + 2as

Distance Travelled by an Object in t th Second:

Let the initial velocity be u, and the object be moving with an acceleration a. Distance traveled by an object in t seconds is S_t = ut + ½ at² 

Distance travelled in (t-1) seconds is S_(t-1) = u (t – 1) + ½ a (t – 1 )²

So, the difference between these two equations gives the distance traveled by the object in t^{th second.}

              S= S_t – S_(t-1)

                = {ut + ½ at²} – {u (t – 1) + ½ a (t-1)²}

           = {ut + ½ at²} – {ut – u + ½ a (t²- 2t + 1)}

          S= ut + ½ a t²- ut + u – ½ a (t² -2t +1)

                 = u+ ½ a t² – ½ a ( t² – 2t + 1 )

                       = u + ½ a (2t – 1)

Equations of Motion Under Gravity

If in the three equations of motion, a is replaced by g and s by h, then equations for the objects falling towards the earth are obtained. Thus, the equations of motion under gravity becomes:

V= u+gt

H=ut + ½ gt²

V²=u²+2gh

If an object is thrown upward away from the earth, then we substitute -g in place of g in these equations.

Equation of Motion Along a Smooth Inclined Plane:

Consider an object O sliding down a smooth inclined plane AB. The acceleration of O is less than g if the object falls freely downwards. The acceleration of the object O sliding down the plane is g sinθ. So, the equations for a body sliding down a smooth inclined plane are:

V = u + (g sinθ) t

S = ut + ½ (g sinθ) t²

V² = u² + 2 (g sinθ) s

If the body moves up along an inclined plane, then its acceleration will be (-g sinθ), and the equations will become:

V = u – (g sin θ) t

S= ut – ½ (g sin θ) t²

V²=u² – 2(g sin θ) s

Some more useful formulas for motion under gravity

When a body is just dropped, u=0

For a body thrown vertically with initial velocity u,

Maximum height = h= u²/2g

Time of ascent= time of descent = u/g

 Total time of flight = 2u/g

The velocity of fall at the point of projection = u