When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion.
There are three such equations. These are:
`v = u + a t` ........(8.5)
`s = u t + ½ a t^2` ........(8.6)
`2 a s = v^2 – u^2` .........(8.7)
where `u` is the initial velocity of the object which moves with uniform acceleration `a` for time `t, v` is the final velocity, and `s` is the distance travelled by the object in time t. Eq. (8.5) describes the velocity-time relation and Eq. (8.6) represents the position-time relation.
Eq. (8.7), which represents the relation between the position and the velocity, can be obtained from Eqs. (8.5) and (8.6) by eliminating t. These three equations can be derived by graphical method.
`ul"EQUATION FOR VELOCITY-TIME RELATION"`
Consider the velocity-time graph of an object that moves under uniform acceleration as
shown in Fig. 8.8 (similar to Fig. 8.6, but now with `u ≠ 0`). From this graph, you can see that initial velocity of the object is u (at point A) and then it increases to v (at point B) in time t.
The velocity changes at a uniform rate a.
In Fig. 8.8, the perpendicular lines BC and BE are drawn from point B on the time and the velocity axes respectively, so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by `OC. BD = BC – CD`, represents the change in velocity in time interval t.
Let us draw AD parallel to OC. From the graph, we observe that
`BC = BD + DC = BD + OA`
Substituting `BC = v` and `OA = u`,
we get `v = BD + u`
or `B D = v – u` (8.8)
From the velocity-time graph (Fig. 8.8), the acceleration of the object is given by
` a = text ( Change in velocity)/text( time taken)`
` = (BD)/(AD) = (BD)/(OC)`
Substituting `OC = t`, we get
` a = (BD)/t`
or `B D = a t` (8.9)
Using Eqs. (8.8) and (8.9) we get
` v = u + a t`
`ul"EQUATION FOR POSITION-TIME RELATION"`
Let us consider that the object has travelled a distance s in time t under uniform acceleration a. In Fig. 8.8, the distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB.
Thus, the distance s travelled by the object is given by
`s =` area `OABC` (which is a trapezium)
`=` area of the rectangle `OADC +` area of the triangle `ABD`
`= OA xx OC + 1/2 (AD xx BD)` (8.10)
Substituting `O A = u, O C = A D = t` and `BD = a t`, we get
` s = u xx t + 1/2 (t xx a t)`
or ` s = ut + 1/2 at^2`
`ul"EQUATION FOR POSITION–VELOCITY RELATION"`
From the velocity-time graph shown in Fig. 8.8, the distance s travelled by the object in time t, moving under uniform acceleration a is given by the area enclosed within the trapezium `OABC` under the graph.
That is, `s =` area of the trapezium `OABC`
` = ( (OA + BC) xx OC)/2`
Substituting `OA = u, BC = v` and `OC = t, ` we get
` s = ( (u + v) t)/2` ........(8.11)
From the velocity-time relation (Eq. 8.6), we get
` t = (v -u )/a` .........(8.12)
Using Eqs. (8.11) and (8.12) we have
` s = ( (v + u) xx (v - u) )/(2a)`
or `2 a s = v^2 – u^2`