Transcription of Motion: Velocity and Net Change
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Motion: Velocity and Net Change
Math 131,applicationsmotion: Velocity and net Change 1 motion : Velocity and Net ChangeIn Calculus I you interpreted the first and second derivatives as Velocity and accel-eration in the context of motion . So let s apply the initial value problem results tomotion problems. Recall that s(t) =position at timet. s (t) =v(t) = Velocity at timet. s (t) =v (t) =a(t) =acceleration at a(t)dt=v(t) +c1= Velocity . v(t)dt=s(t) +c2=position at will need to use additional information to evaluate the that the acceleration of an object is given bya(t) =2 costfort 0 with v(0) =1, this is also denotedv0 s(0) =3, this is also (t). findv(t)which is the antiderivative ofa(t).
math 131, applications motion: velocity and net change 2 Displacement vs Distance Travelled DEFINITION 6.1. The displacement of an object between times t = a and a later time t = b is s(b) s(a) = Zb a v(t)dt. In other words, displacement is the net area between the velocity curve and the horizontal axis, while the distance travelled is the total area between the ...
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