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Derive the kinematic equations of motion for constant acceleration . |
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Answer» Solution :Let US consider an object moving in a STRAIGHT line with uniform or constant acceleration 'a' Let u be the velocity of the object at time t = 0 Let v be velocity of the body at a later time t. Velocity - time relation :The acceleration of the body at any instant is given by ` = (dv)/(dt) or dv = a dt` Integrating both SIDES with the condition that as time CHANGES from 0 to t, the velocity changes from u to v. For the constant acceleration, `underset(u)overset(v)intdv = underset(0)overset(t)intadt = a underset(0)overset(t)int dt rArr [v]_(u)^(v)a [t]_(0)^(t)` `v - u = at (or) v = u + at "". . . (1)` Displacement-time relation : The velocity of the body is given by `v = (ds)/(dt) or ds = vdt` But , v = u + at, `:. `ds = (u + at) dt Let us assume that initially at time t=0, the particle started from the origin. At a later time t, the particle displacement is s. Further assuming that acceleration is time - independent, we get `underset(0)overset(z)int ds = underset(0) overset(t)int u dt = underset(0)overset(t) int at dt (or)` `:. s = ut +(1)/(2) at^(2)`. . . (2) Velocity- displacement . relation :The acceleration is given by the first derivative of velocity with respect to time . `a = (dv)/(dt) = (dv)/(ds)(ds)/(dt) = (dv)/(ds) v` Let is displacement traversed [since ds/dt = v] `:. a = (1)/(2) (dv^(2))/(ds)` or ds ` = (1)/(2a) d (v^(2))` When the velocity changes from `u^(2) "to"v^(2)` , displacement changes from 0 to s, integrating the above EQUATION, we get `underset(0)overset(s)int ds = underset(m) overset(v) int (1)/(2a) d(v^(2))` `:. s = (1)/(2a) (v^(2) - u^(2))` `rArr :. v^(2) = u^(2) + 2as "". . . (3)` Let us derive the displacement s in terms of initial velocity u and finla velocity v . From the equation (1) we can write, at = v - u Substitute this in equation (2), we get `s = ut (1)/(2) (v -u) t` `:. s = ((u +v)t)/(2)` |
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