function s(t) = t/( t + 1) describes the distance (in feet) of an object from a specific location P along a line l at the end of t seconds where t ≥ 0

Question

Find the velocity (in feet per second) of the object at t = 6 seconds.

Answer 1

s(t) = t/(t+1)

velocity = the derivative of the distance

v(t) = s'(t) = -t/(t+1)^2 + 1/(t+1)

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

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

v(6) =-(6^2+6)/(7^2)

velocity after 6 seconds

=-42/49= -6/7 feet per second

it's going backwards at about 0.857 ft/sec