The propellers of a WWII destroyer produced a thrust of 2.65MN to drive the ship
ID: 2257237 • Letter: T
Question
The propellers of a WWII destroyer produced a thrust of 2.65MN to drive the ship at a top speed of 17m/s. If the negines were suddenly thrown into reverse, developing one half the thurst they produced driving ahead, (A) for how many seconds would the ship travel before being brought to a stop, neglecting the drag force of the water? the mass of the ship is 2.0*10^6kg. (B) assume the drag force from the water is linearly proportional to speed. Now what is the time required to bring the ship to a stop? (adding drag should cut stopping time significantly)
For part A) my answer is:
t=25.66s
For B) I have found this solution but dont understand how they jump from the bolded sections (the steps from getting from the left of the red arrow to the right side of the red arrow) can someone help me understand this or provide a alternative solution for part B)?? THANKS!
b)now drag force,Fd = kv
since at v = 17 m/s,Fd = 2.65*10^6 N, this gives k = 1.56*10^5
new force,F" = -1.325*10^6 - kv N
a = F"/m = d^2x/dt^2 = dv/dt = [-1.325*10^6 - kv]/m
=>dv/[1.325*10^6 + kv] = -dt/m => dv/1.56*10^5[8.49 + v] = -dt/m => dv/[v+8.49] = -0.078dt
integrating both sides limits of v being from 17 to 0 and time being from 0 to t
ln[v+8.49]|170 = -0.078t |0t
ln[8.49/(8.49+17)] = 0-t => -t = ln[8.49/25.49] => t = 1.0994 s Ans(b)
Explanation / Answer
dv/dt = [-1.325*10^6 - kv]/m
by rearranging terms-
dv/[1.325*10^6 + kv] = -dt/m
=> dv/[1.325*10^6 + 1.56*10^5v] = -dt/m
Taking k = 1.56*10^5 common in Dr
dv/1.56*10^5[8.49 + v] = -dt/m
=> dv/1.56*10^5[8.49 + v] = -dt/ 2.0*10^6
=> dv/ [8.49 + v] = -1.56*10^5 * dt/ 2.0*10^6
=> dv/[v+8.49] = -0.078dt
=>
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