The propellers of a WWII destroyer produced a thrust of 2.65 MN to drive the shi
ID: 2218415 • Letter: T
Question
The propellers of a WWII destroyer produced a thrust of 2.65 MN to drive the ship at a top speed of 17 m/s. If the engines were suddenly thrown into reverse, developing one-half the thrust they produced driving ahead, 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 x 10^6 kg. (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?Explanation / Answer
Force,F = 2.65*10^6 N velocity,v = 17 m/s i.e at 17 m/s drag equals thrust new force ,F' = -F/2 = -1.325*10^6 N initial velocity,u = 17 m/s mass m = 2*10^6 kg since no drag force acceleration ,a= F'/m = -0.6625 m/s final velocity ,v= 0 v = u + at 0 = 17 - 0.6625t => t = 25.66 s Ans(a) 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)
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