Problem #5 #5A. A plane leaves Seattle, flies 85 mi at 22o north of east, and th

Question

Problem #5

#5A. A plane leaves Seattle, flies 85 mi at 22o north of east, and then changes direction to 48o south of east. After flying 115 mi in this new direction, the pilot must make an emergency landing on a field. The Seattle airport facility dispatches a rescue crew.
a) In what direction and how far should the crew fly to go directly to the field? Use the method of components to solve.
b) Use graphical analysis (ruler and protractor to scale) to determine the answer to part a.
#5B. Relative to the ground, a car has a velocity of 18.0 m/s, directed due north. Relative to this car, a truck has a velocity of 22.8 m/s, directed 52.1o south of east. Find the magnitude and direction of the truck's velocity relative to the ground. Make sure you draw vector addition graphs to help show what you are doing.

Explanation / Answer

5A. Call due east zero degrees Step One ======= Write your vectors using east as 0o Vector A = 85 miles @ 22o Vector B = 115 miles @ - 48o = 360 -48 = 312o Step Two ======= Calculate the the two Horizontal components and then add them. A_h = 85*Cos(22) miles A_h =. . . . . . . . . . . . . . . . . 78.81 miles B_h = 115Cos(312) miles B_h = . . . . . . . . . . . . . . . . . 75.95 miles Total . . . . . . . . . . . . . . . . . 158.6 miles total Step Two ======= Calculate the the two Vertical components and then add them. A_v = 85*sin(22) miles A_v =. . . . . . . . . . . . . . . . . 31.84 miles B_v = 115*sin(312) miles B_v = . . . . . . . . . . . . . . . . - 85.46 miles Total . . . . . . . . . . . . . . . . . - 53.62 miles total Comment ======== The horizontal is positive, the vertical negative; the angle must be in quadrant 4. Step 3 ===== Calculate the angle as though it were in quadrant 1 (which means ignore the signs.) Tan(x) = vertical component total / horizontal component total. Tan(x) = 53.62/ 158.6 Tan(x) = 0.338 x = tan(-1) [0.496] x = 18.67 o Since this must be in the 4th quadrant, it just so happens to be the answer without adjustment. Answer: 26.4 degrees south of east. Step 5 ===== Find the distance. d = sqrt(horizontal total^2 + vertical total ^2) d = sqrt( 158.6^2 + 53.62^2 ) d = 167.41 miles from the starting point. 5B. V of T in relation to G (Vtg) = V of T in relation to C (Vtc) + V of C in relation to G Vcg. The x factor looks like this: Vtgx = Vtc + Vcg 22.8 cos 52.1 + 0 = 13.514 Y factor: Vtgy = Vtc + Vcg -22.8 sin 52.1 = 8.88*10^-3 Vtg = Sqrt (Vtgx^2 + Vtgy^2) = 13.514 m/s Direction: tan^-1 y/x=0.04 dgrees

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