Question

A 5.0-g marble is released from rest in the deep end of a swimming pool. An underwater video reveals that its terminal speed in the water is 0.30 m/s. (a) What is the acceleration of the marble at the instant it is released? (b) What is the acceleration of the marble when it has reached its terminal speed? (c) How long does it take the marble to reach half its terminal speed?

Answer 1

Answer:

a) a = -g = 9.8 m/s² , b) a = 0 m/s² and c)   t1 = 0.0213 s

Explanation:

a) At the moment the marble is released its velocity is zero, so it has no resistance force, the only force acting is its weight, so the acceleration is the acceleration of gravity

       a = -g = 9.8 m / s²

b) When the marble goes its terminal velocity all forces have been equalized, therefore, the sum of them is zero and consequently if acceleration is also zero

      a = 0 m / s²

c) We have to assume a specific type of resistive force, for liquid in general the resistive force is proportional to the speed of the body.

The expression of this situation is

         v = mg / b (1 -$e^{-bt/m}$ )

For a very long time the exponential is zero, so the terminal velocity is

        $v_{T}$ = mg / b

        b = mg /  $v_{T}$

        b = 5 10-3 9.8 / 0.3

        b = 0.163

We already have all the data to calculate the time for v = ½ $v_{T}$

        ½ $v_{T}$ = $v_{T}$ (1 -$e^{-bt/m}$)

        ½ = 1- e (- 0.163 t1 / 5 10-3)

        e (-32.6 t1) = 1-0.5              (by  ln())

       -32.6 t1 = ln 0.5

       t1 = -1 / 32.6 (-0.693)

       t1 = 0.0213 s