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Derive the kinematic equations of motion for constant acceleration. |
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Answer» Solution :Consider an object moving in a straight line with uniform or constant acceleration `alpha`. Let u be the velocity of the object at time t = 0, and v be velocity of the body at a later time t. Velocity - time relation: (i) he acceleration of the body at any instant is given by the first derivative of the velocity with respect to time, `a=(dv)/(dt)ordv=adt` INTEGRATING both sides with the condition that as time changes from 0 to 1, the velocity changes from u to v. For the constant acceleration, `int_(u)^(v)dv=int_(0)^(t)adtimplies[v]_(u)^(v)=a[t]_(0)^(t)` `v-u=at(or)v=u+at` Displacement - time relation (ii) The velocity of the body is given by the first drivative of the displacement with respect to time. `v=(ds)/(dt)ORDS=vdt` and since v=u+at, We GET ds =(u+at)dt Assume that initially at time 1 = 0, the particle started from the origin. At a later time i, the particle displacement is s. Further assuming that acceleration is time-independent, we have `int_(0)^(t)ds=int_(0)^(t)udt+int_(0)^(t)atdtors=ut+(1)/(2)at^(2)""......(2)` Velocity - displacement relation (iii) The acceleration is given by the first derivative of velocity with respect to time. `a=(dv)/(dt)=(dv)/(ds)(ds)/(dt)=(dv)/(ds)v` [since ds//dt=v] where s is displacement traversed. This is rewritten as `a=(1)/(2)(d)/(ds)(v^(2))ords=(1)/(2a)d(v^(2))` Integrating the above equation, using the fact when the velocity changes from u to v, displacement changes from 0 to s, we get `int_(0)^(t)ds=(v_(1))/(""^(=)2a)d(v^(2))` `:.s=(1)/(2a)(v^(2)-u^(2))` `:.v^(2)=u^(2)+2as""....(3)` We can also derive the displacement s in terms of initial velocity u and final velocity v. From equation we can write, at=v-u SUBSTITUTE this in equation, we get `s=ut+(1)/(2)(v-u)t` `s=((u+v)t)/(2)"".......(4)` The equations are called kinematic equations of motion. |
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