Where e is the base of the natural logarithm. How much time has passed when the

Question

Where e is the base of the natural logarithm. How much time has passed when the submarine's speed reaches 59% of its terminal value? What is the magnitude of the submarine's acceleration at this time? A toy submarine of mass m 0.93 kg moves around a submerged circular track of radius R = 1.72 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 isC = 1.21 kgs. Assuming it starts from rest at t = 0 s, the speed of the submarine at a later time t is given by where e is the base of the natural logarithm. How much time has passed when the submarine's speed reaches 59% of its terminal value? Number sec What is the magnitude of the submarine's acceleration at this time? Number m/s

Explanation / Answer

at time t = 0, terminal speed V = F/C

time taked to reach 59% of terminal speed is

F/C *(1 - e^-Ct/m) = 0.59 F/C

e^-Ct/m = 1- 0.59

-Ct/m = ln (0.41) = -0.891

t = 0.891 * m/C

t = 0.891 * 0.93/(1.21)

t = 0.684 secs

the accleration a of the submarine a(t) = dV/dt

a = d(F/C * (1 - e^-Ct/m)/dt

a = -F/C e^-Ct/m *(-C/m)

a = F/m * e^-Ct/m

a = (4.98/0.93) * (e^-(1.21* 0.684/0.93)

a = 2.2 m/s^2

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