Question

A particle moves falling vertically through a viscous liquid. The velocity is recorded as a function of time and it is observed that it corresponds approximately to the following function: v (t) = 10 (1 - exp (-t)), t in seconds v in m / s. a) Graph velocity versus time. Describe what happens in words. b) Calculate the instantaneous acceleration of the particle and subtract from it the acceleration of gravity (10 m / s²). Graph the resulting acceleration, which can be said to correspond to viscous braking in the fluid. c) Find some relationship between viscous braking and speed. d) calculate the position versus time of the particle and draw the graph. Show procedure

Answer 1

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(a) Graph velocit of v versus t OLIO 0.366-4 2 1.36 8.6 do o 10 time. V = 10c1-et is the instantaneous velout = 10(1-9) as tinereases, reachey steady value Que to viscous force, pantuele reaches to uniform velocity called terminal velocity o 24 aceleration abt) - duct =d[ioci-et] dt at =10 core-t) alt) = loet As time mureases, a tend to zero. Le velocity becomes steady cuniformie particle is in its ↑ acecleration retardi 29 motor time. 2. Assuming from viscous braking > acceleration, nexox v= Loci-e-t = 10 - 10 e- ie 10 = 1o-a Position time