Question

A boat going down river is initially traveling at 25 knots (1 knot 1.151 mph). Determine the distance traveled past starting point A in 5 minutes if the boat travels at 25 knots relative to the shore for 1 minute before cutting its engines. The river current flows at 2 knots causing the boat's velocity relative to the shore to follow the function v = 42.2-6in(t- 59) ft/s after the engines are cut, where is in seconds. v (ft/s) 40 2041 2 t (s) 100 300 500 Hint: lln(x)dx = xln(x)-X

Answer 1

Imph = 1.467 tis. initial velocity a u= 25 x 10 1514 10 467 = 42,213 ft/s distance travelled before engines are cut ft di = uxt = 42.213 x Go =2532.78 dt Distance travelled after engines are cut. we know that velocity = v= dx 5 . Vd: integrating both aside dx = luat =642.2 - 6 144 - 599) dt 5 X 60 O GO t = 300 ac = - 6 300 I inlt-59) dt 42.2 d+ t = 60 300 **42.2 [+ 5.- 6 [e-salt-s)-4). 3) *= 42,2/200-60) - 6/200-59)/(200-59) – zoolat -{(60-59) L (60-59) -60%)

* = 42, 240 240 - 68241 10 241 - 300? - &1601-60}] - 10128 - 65 241x 5.848 - 300 -0 +60) 310128 - G[ 1321.83 – 240] = 10128 - 6491,02 x= 3636.98 ft] Total distance D= d+n D=2532.78+ 3636.98 (D = 6 169.76 ft * | 1n (4659) dt = 6 (4-59) fot - t (104-59) • Jt oot - G ( 59) - số di - tlon (+-59) - J + - 59 that ( 53 dt It-59 dt t-59 t-59 = 4 ln (+-59) - t - 59 ln (+ -59) (t - 59)dt = (t - 59) Colt -59) nt

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