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Water traveling along a straight portion of a river normally flows fastest in the middle, and the speed slows to almost zero at the banks.

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Water traveling along a straight portion of a river normally flows fastest in the middle, and the speed slows to almost zero at the banks. Consider a long straight stretch of river flowing north, with parallel banks 100 m apart. If the maximum water speed is 3 m/s, we ca function as a basic model for the rate of water flow x units from the west bank: f(x) = -x(100 - x). 2500 (a) A boat proceeds at a constant speed of 5 m/s from a point A on the west bank while maintaining a heading perpendicular to the bank. How far down the river on the opposite bank will the boat touch shore? X Your answer cannot be understood or graded. More Information m Graph the path of the boat. y 50 20 40 10 30 X 20 40 60 80 100 20 -10 10 -20 - X O 20 40 60 80 100 y 50 20 40 10 30 X 20 20 40 60 80 100 -10 10 O-20 LX 20 40 60 80 O 100 (b) Suppose we would like to pilot the boat to land at the point B on the east bank directly opposite A. If we maintain a constant speed of 5 m/s and a constant heading, find the angle at which the boat should head. (Round your answer to one decimal place.) south of east Graph the actual path the boat follows. y 50 50 40 40 30 30 20 20 10 10 X O 20 40 60 80 LX 100 20 40 60 80 100 4 y 20 20 10 - X - X 20 40 60 80 100 20 40 60 80 100 -10 10 O-20 O-20 Homework 13.4 Motion in Space - MTH 265 8-Week Summer 2021, section E56M SU21, Summer 1 2021 |EXAMPLE 6 A projectile is fired with muzzle speed 130 m/s and an angle of elevation 45 from a position 30 rn above ground level. Where does the projectile hit the ground and with what speed? SOLUTION If we place the origin at ground level, then the initial position of the projectile is (0, 30) and so we need to adjust the parametric equations of the trajectory by adding 30 to the expression for y. With v0 = 130 m/s, at = 45, and g = 9.8 m/sz, we have x = 130 cos(1r/4)t = y = so + 130 sin(11:/4)t 7 i913)? : Impact occurs when y = 0, that is 4.9!:2 7 65t 7 30 = 0. Solving this quadratic equation (and using only the positive value of t), we get the following. (Round your answer to two decimal places.) t= ss + (8,450 + 588 ~|:| 9.8 Then x :6 65(19.08) : E (rounded to the nearest whole number), so the projectile hits the ground about :I m away. The velocity of the projectile is V(t) : WU\") : So its speed at impact (rounded to the nearest whole number) is lv(19.08)| = \\/(65\\/E)2 + (am/E 9.8-19.08)2 x mfs. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER EXAMPLE 3 A moving particle starts at an initial position r(0) 2

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