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

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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?

Physics

1 answer:

serg [2.5K] · Answer 1 · 2026-03-19T18:12:58+03:00

a) The marble's acceleration when released is a = -g = 9.8 m/s²; b) When it reaches terminal velocity, the acceleration is a = 0 m/s²; c) The time taken to reach half of its terminal speed is t1 = 0.0213 s. When the marble is dropped, its velocity is zero, meaning there is no resistive force acting on it; the only force present is gravity, resulting in an acceleration equivalent to gravity's rate: a = -g = 9.8 m/s² At terminal velocity, all forces balance out, which leads to zero net acceleration: a = 0 m/s² Assuming a specific resistive force for a liquid, this force is generally proportional to the object's velocity. The relevant equation for this scenario is: v = mg/b (1 - e^(-bt/m)) Long term, the exponential approaches zero, so terminal velocity is expressed as: v_terminal = mg/b and, subsequently, b = mg/v_terminal Substituting gives: b = 5 x 10^-3 kg × 9.8 m/s² / 0.3 m/s = 0.163 Using this data, we can calculate the time to reach v = 0.5 × v_terminal: 0.5 × v_terminal = mg/b (1 - e^(-bt/m)) Thus, 0.5 = 1 - e^(-0.163t1/(5 x 10^-3)) e^(-32.6t1) = 0.5 Taking the natural logarithm gives: -32.6t1 = ln(0.5) t1 = -1/(32.6)(-0.693) = 0.0213 s.