Uniformly Accelerated Motion

We all know that acceleration is the rate at which the velocity of an object changes with respect to time. However, do you know what is uniformly accelerated motion? It can be difficult to perceive how an aspect that is defined by a rate of change can be defined as uniform. Therefore, let us study such uniformity in acceleration in detail, especially since this a vital concept that is necessary for numerous other chapters too.

Define Uniform Acceleration

One can say that uniform acceleration definition refers to an acceleration of an object, which remains constant irrespective of time. In simpler terms, a number equal to the acceleration in such a motion does not change as a function of time.

Some uniformly accelerated motion examples include a ball rolling down a slope, a skydiver jumping out of a plane, a ball dropped from the top of a ladder and a bicycle whose brakes have been engaged.

Keep in mind that these examples of uniform application do not maintain absolute uniformity of acceleration, due to the interference of gravity and/or friction. However, these are still some of the cases where acceleration would be uniform if gravitational force and friction is considered zero.

Equations of Uniformly Accelerated Motion

After understanding what is uniform acceleration, one should proceed to learn about the three kinematic equations that define such a motion. 

1. Velocity Equation

To understand the first equation, refer to the graph below.|

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Let us assume that the initial velocity of an object is u. Similarly, uniform acceleration for this object is a. Now, the object reaches at point B after time t, post which, its velocity becomes v.

DA represents a parallel line to the x-axis from the point where the object started moving. BA is a line parallel to the y-axis which is connecting A to the end-point of the body. This vertical line meets the x-axis at point E. Consider OD as u.

Suppose OE = t or time

From this graph, we can derive

BE = AB + AE

v = DC + OD

From the graph, it is apparent that DC = AB and OD = AE (OD is also u)

v = DC + u (i)

Now, we know acceleration (a) = \[\frac {(v-u)} {t}\]

Thus, a = \[\frac {OC-OD} {t}\]

a = \[\frac {DC} {t}\]

at = DC   (ii)

Substituting the value of DC from (ii) to (i)

V = at + u

This is the first kinematic equation for uniformly accelerated motion.

2. Distance Equation

The key to define uniform acceleration is the second equation to determine distance. 

Distance (s) = Area of ABD + Area of ADOE

s = \[\frac {1} {2}\] x AB x AD + (OD x OE)

s = \[\frac {1} {2}\] x DC x AD + (u + t)

AB = DC

s = \[\frac {1} {2}\] x at x t + ut

s = \[\frac {1} {2}\] at² + ut

or, s = ut + \[\frac {1} {2}\] at²

3. Equation to Relate Distance and Velocity

The third and final equation to define uniform acceleration is the one that relates distance (s) and velocity (v). 

Area of trapezium ABDOE = ½ x (sum of parallel sides – distance between parallel sides)

s = \[\frac {1} {2}\] (DO + BE) x OE

Therefore, s = \[\frac {1} {2}\](v + u) x t …….(iii)

From equation (ii), we know a = \[\frac {(v-u)} {t}\]

t = \[\frac {(v-u)} {a}\] …….(iv)

Therefore, s = \[\frac {1} {2}\] (v + u) x (v – u)/a

s = \[\frac {1} {2}\]a (v + u)(v – u)

2as = v² – u²

v² = u² + 2as

Thus, this is how we write the three equations of uniformly accelerated motion.