and initial velocities v and v of an object moving with uniform acceleration a : v = v + at ( . ) This relation is graphically represented in Fig. . .
The area under this curve is : Area between instants and t = Area of triangle ABC + Area of rectangle OACD v v t + v t Fig. . Area under v-t curve for an object with uniform acceleration . As explained in the previous section, the area under v-t curve represents the displacement.
Therefore, the displacement x of the object is : v v t + v t ( . ) But a t Therefore, a t + v t or, ( . ) Equation ( . ) can also be written as v + v t v t ( .7a) where, (constant acceleration only) ( .7b) Equations ( .7a) and ( .7b) mean that the object has undergone displacement x with an average velocity equal to the arithmetic average of the initial and final velocities.
From Eq. ( . ), t = ( v – v ) /a . Substituting this in Eq.
( .7a), we get = ax ( . ) This equation can also be obtained by substituting the value of t from Eq. ( . ) into Eq.
( . ). Thus, we have obtained three important equations : ax ( .9a) connecting five quantities v , v, a, t and x . These are kinematic equations of rectilinear motion for constant acceleration.
The set of Eq. ( .9a) were obtained by assuming that at t = , the position of the particle, x is . We can obtain a more general equation if we take the position coordinate at t = as non- zero, say x . Then Eqs.
( .9a) are modified (replacing x by x – x ) to : ( .9b) ( a x ( .9c) Example . Obtain equations of motion for constant acceleration using method of calculus. Answer By definition d v = a d t Integrating both sides