A particle moves along the x-axis, its position given by x(t) = t/(1 + t^2), t >= 0, where t is measured in seconds and x in meters.

Question

A particle moves along the x-axis, its position given by , where t is measured in seconds and x in meters.

(a) Find the acceleration at time t.

(b) When is the particle speeding up?

(c) When is the particle slowing down?

(a) a(t) = -(2t(-t^2+3))/((t^2+1)^3)

(b) ___ < t < ___

(c) ___ < t < ___ and t > ___

Answer 1

take the 2nd derivatve

a(t)= (2t^3-6t)/(t^2+1)^3

it speeds up when t>sqr3 when t> about 1.7 seconds

it slows down when 0<t<sqr3 when t < 1.7 seconds, and positive

set the numerator of a(t) = 0 and solve for t

2t(t^2-3) = 0

t^2 = 3

t = sqr3 (ignore negative square root if t=/> 0)

when a(t)=0, it is stationary, for an instant, neither speeding up or down