Ask a New Question
Menu

Bubba decides to go fishing one day. He gets into his boat and starts out from the boat ramp to find the perfect fishing spot in the lake. First, he goes 121 meters in a direction of 20° west of north. Then, he goes 104 meters in a direction of 33.0°south of west. He turns and goes 211

meters in a direction of 20.5°south of east. Finally, he turns and heads 135 meters in a direction of 41.8°north of east. He then realizes that he forgot his fishing pole.

What direction and distance must he go in order to return directly to the boat ramp?

If he were moving at 1.55 m/s the entire time, how long did it take him to reach the spot where he realized that he had forgotten his fishing pole?

If he heads back to the boat ramp on the most direct path at the same speed of 1.55 m/s, how long does it take him to return to the boat ramp?

4 answers

add all your north distances
for example the first one is 121 cos 20
call the sum Y

add all your east distances
for example the first one is -121 sin 20
call the sum X
(note negative east because west)
then your location is (X,Y) when you turn to go back
The distance straight back is sqrt(X^2+Y^2)
the direction from the origin to (X,Y) is found as follows
Tan^-1 (X/Y) = angle of location EAST OF NORTH (watch out what quadrant you are in)
you want to reverse that to go back to the dock so
Tan^-1(X/Y) + 180 is the direction to sail.

the time heading out is (121 + 104 + ..... )/ (1.55)
the time back is sqrt(X^2+Y^2)/(1.55)
1. All angles are measured CCW from +X-axis.
Displacement = 121m[110o] + 104m[213o] + 211m[339.5o] + 135m[41.8o].
X = 121*Cos110 + 104*Cos213 + 211*Cos339.5 + 135*Cos41.8 =
Y = 121*sin110 + 104*sin213 + 211*sin339.5 + 135*sin41.8 =
Magnitude = Sqrt(X^2 + Y^2) =
Direction = Arctan(Y/X) =
X = 169.7 m.
Y = 73.15 m.
Magnitude = 184.8 m.
Direction = 23.3o N. of E. = 23.3o CCW.
2. d = r * t.
t = d/r = (121+104+211+135)/1.55 = Time in seconds.

3. D = r * t.
t = D/r = 184.8/1.55 = Time in seconds.