2-3 KinematicalequationsDefinition of v, a.Differential equationsd(t)a(t)d(t) = adtdtdr(t)' dv(t) = f'adt(t)?dtIntegral equationsv(t) - v(to)= ['adt(t)=v(to)+ I a(t)dtr(0)=r(0)+ I' (i)dt(0)=l=-0Initial condition福(t) = f(t)+CFinding C.V, =v,(to)+fa,idt x=x(to)+ f'vedt
2-3 Kinematical equations Definition of v, a. ( ) ( ) ( ) ( ) v t dt dr t a t dt dv t = = Integral equations = + = + t t t t r t r t v t dt v t v t a t dt 0 0 ' ' 0 0 ( ) ( ) ( ) ( ) ( ) ( ) v t f t C ( ) ( ) = + Initial condition Finding C. 0 ( ) 0 t t v t v = = ' ' 0 0 v v (t ) a t dt t t x = x + x dv t adt ( ) = = t t t t dv t adt 0 0 ( ) − = t t v t v t adt 0 ( ) ( ) 0 = + t t x x t vx dt 0 ( ) 0 Differential equations
ExampleFinding velocity from Acceleration。 =(0,5,-3)m / s, a =(0,0,-2)m / s2Find its velocity after5s(t)= Vo + [ adt = V + [ adtv, =vox +fa,(t)di =0+ ['odtV, = 5m/ sv, = Vo +[a,(t)dt =-3+[(-2)dt =-3-2x5= -13(m / s)
Finding velocity from Acceleration (0,5, 3) / , 0 v = − m s 2 a = (0,0,−2)m/s = + = + 5 0 0 0 0 v(t) v adt v adt t t = + = + t t x x x v v a t dt dt 0 ' 0 ' ' 0 ( ) 0 0 v m s y = 5 / Example: Find its velocity after 5s. ( ) 3 ( 2) 3 2 5 5 0 ' 0 ' ' = 0 + = − + − = − − v v a t dt dt t z z z = −13(m/s)