Equations On Uniformly Accelerated Motion

A body is in a continuous motion when it moves with uniform acceleration i.e it undergoes equal change in velocity in equal intervals of time. If the body is moving in a uniform manner that means there is a fixed external force acting on the body.

What is uniform acceleration?

A body with uniform acceleration moves with constant speed. This happens when the same external force is applied to the body. 

For instance, if a car is moving at a speed of 60 km/hour for 2 hours that means the acceleration of the car in 2 hours is zero. 

What are the equations on the uniformly accelerated motion? 

Let us derive the equations on uniformly accelerated motion

Let the initial velocity be u, at which the car started and reach the final velocity (v). The time t is required for getting the final velocity.  The total distance s covered over the period of time. All these parameters are connected by three equations on the uniformly accelerated motion:

Velocity – time relationship 

Let us take initial time as zero, final velocity as v, initial velocity as u and final time as t, we get

We know that the equation a = (v –u ) / t)

Rearranging this equation, we can get

                        a t = v – u

    Or              v = u + at ———-(1)

It is the first equation of motion.

Position – time relation

If the final velocity of a body is v’ and the initial is v, then its average velocity can be mentioned as

                        vav= (1/2) (v + v’) ———– (2)

Suppose initially (at time = t), the position of the body is x. At x’, the average velocity can be calculated as:

vav  = (x’ – x )/ (t’ – t ) (t’ is the time when the object is at x’)

Or  x’- x  = (t’ – t) vav  ————————- (3)

We will use the value of vav from (2) into (3)

            x’ – x = (1/2) (t’ – t) (v + v’) ———— (4)

From equation (1),  we can say that v’ = v + a (t’ – t)

Using this answer in equation (4)

x’ – x =  (1/2) (t’ – t) [v + v +a (t’ – t)]

=  (1/2) (t’ – t) 2v +(1/2)  a (t’ – t) 2

= v (t’ – t ) + (1/2) a (t’ – t )2 ——— (5)

If the initial time is zero and the initial velocity is u, the displacement S for x’ – x and final time as t, then

s = ut + (1/2)at2

It is the second equation of motion

Velocity – displacement relation:

Equation (5) gives x’ – x = v (t’ – t) + (1/2) a (t’ – t)2

We have to use velocity instead of time.

So using a = (v’ – v)/(t’ – t) 

we get t’ – t =(v’ – v) /a

Putting this value in equation (5), we get

x’ – x = v (v’ – v)/(a) + (1/2) a{ v’ – v)/(a)}2

 x’ – x =(v’2 – v2) / 2a              

Or v’2 – v = 2a (x’ – x)

If v is the final velocity and us as the initial velocity, x’ – x is the displacement S, 

then v2 – u2 = 2as 

=>  v2  = u2 + 2as 

It is the 3rd equation of motion.

Displacement in the nth second

Suppose the body moves at a distance S in n seconds

sn-1 is in n – 1 seconds. 

Then using s = ut + (1/2) at2 we get,

sn = un + (½)an2  & sn-1 = u (n – 1) + (1/2)  (n – 1)2

So distance travelled in nth second = sn – sn-1

sn – sn-1 = un + (1/2) an2 – {u (n – 1) + (1/2) a (n – 1)2}

snth  = un + (1/2) an2 – un + u – (1/2) a (n2 + 1 – 2n)

     = (1/2) an2 + u – (1/2) an2 -(1/2) a +(1/2)  a (2n) = u + an – (1/2) a

 snth  = u + (1/2)  a (2n – 1 )

Conclusion

In uniform accelerated motion, the velocity of an object does not change with respect to time. A body is in a continuous motion when it moves with uniform acceleration i.e it undergoes an equal change in velocity in equal intervals of time. Equations of motion of uniform acceleration motion provide the relation between initial velocity, final velocity, acceleration, time, and the distance travel by the object.

For uniformly acceleration motion, equation of motion are following:

s = ½ (v + u)t   —— (1)

v = u + at —– (2)

v2 = u2 + 2as —- (3)

s = ut + 1/2at2   ——- (4)