Choose a polynomial a(t) of any order any with arbitrary coefficients to represe

Question

Choose a polynomial a(t) of any order any with arbitrary coefficients to represent the instantaneous acceleration of an object moving along a straight line. Also pick up its initial position (in meters) and initial velocity (in meters/seconds).
(a) Determine the analytical expression for the instantaneous velocity of the object;

(b) Determine the analytical expression for the position of the object;
(c) Select any time instant and describe the motion of the object at that instant specifying whether it moves in the positive/negative direction, or is at rest, also whether it speeds up/slows down or moves with constant velocity.

Explanation / Answer

a) suppose a(t) = a1.t^x

and initial velocity v(o) = v0

initial position r(0) = r0

as we know dv/dt = a(t) = a1.t^x

dv = a(t).dt = a1.t^x .dt

integrating we get ,

v(t) = a1.t^(x-1) / (x-1)   + a2

V(0) = a2 =v0

s0 , v(t) = a1.t^(x-1) / (x-1)   + v(0)

b)

as we know dr/dt = v(t) = a1.t^(x-1) / (x-1)   + v(0)

dr = v(t).dt = [a1.t^(x-1) / (x-1)   + v(0) ].dt

integrating we get ,

r(t) = a1.t^(x-2) / (x-2)(x-1)   + v0.t   + a3

r(0) = a3 =r0

s0 ,r(t) = a1.t^(x-2) / (x-2)(x-1)   + v0.t   + r0

c)

calculate v(t) : if zer0 then at rest

+ve ...moving in+ve direction

-ve ...moving in -vs direction

diffrenciate the v(t)

dv(t) /dt

if dv(t) /dt is zer0.   .....speed is constant.

dv(t)/dt is +ve ....speeding up

dv(t)/dt is -ve .....speeding down

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