A toy submarine of mass m = 0.63 kg moves around a submerged circular track of r

Question

A toy submarine of mass m = 0.63 kg moves around a submerged circular track of radius R = 2.02 m. The submarine's engine provides a constant propulsion force of F = 4.98 N. When the sub is in motion, it is subject to a viscous drag force exerted by the water. This force is proportional to the sub's speed; the proportionality factor is C = 1.31 kg/s. Assuming it starts from rest at t = 0 s. How much time has passed when the submarine's speed reaches 63% of its terminal value? What is the magnitude of the submarine's acceleration at this time?

Question 18 of 18 Map sapling learning Oy submarine of mass m = 0.63 kg moves around a submerged circular track of radius R= 2.02 m ne submarine's engine provides a constant propulsion force of F= 4.98 N. When the sub is in motion, it is subject to a viscous drag force exerted by the water. This force is proportional to the sub's speed; the proportionality factor is C= 1.31 kg/s. Assuming it starts from rest at t= 0 s, the speed of the submarine at a later time t is given by - Ctl m where e is the base of the natural logarithm How much time has passed when the submarine's speed reaches 63% of its terminal value? Number sec What is the magnitude of the submarine'sS acceleration at this time? Number m/ s Previous Give Up & View Solution O Check Answer > Next Exit Hint

Explanation / Answer

v(t) =F/C ( 1 - e-Ct/m)

we need to find the time when v(t) = 0.63F/C

so, 0.63F/C = F/C ( 1 - e-Ct/m)

so e-Ct/m = 1-0.63 = 0.37

or -Ct/m = ln 0.37 = -0.99

t = 0.99 * 0.63 / 1.31 = 0.47secs

2) speed at t = 0.47secs ; v(t) = F/C(1- e-1.31*0.47/0.63) = 4.98/1.31(1- e-1.31*0.47/0.63) = 3.8 (1-0.376) =2.37m/s

since the submarine is moving along a circular path, so

ma = mv2/R

from here, a = v2/R = (2.37)2/ 2.02 = 2.78m/s2

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