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?
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.
To find the direction and distance for Bubba to return directly to the boat ramp, we can use vector addition. We need to add up all the individual displacements to get the net displacement, which will give us the direction and distance.
First, let's break down the given displacements into their x and y components by using trigonometric equations.
1. 121 meters at 20° west of north:
X component = 121 * sin(20°)
Y component = 121 * cos(20°)
2. 104 meters at 33.0° south of west:
X component = -104 * cos(33.0°)
Y component = -104 * sin(33.0°)
3. 211 meters at 20.5° south of east:
X component = 211 * cos(20.5°)
Y component = -211 * sin(20.5°)
4. 135 meters at 41.8° north of east:
X component = 135 * cos(41.8°)
Y component = 135 * sin(41.8°)
Now, we can find the net displacement by summing up the x and y components:
Net X component = Sum of all X components
Net Y component = Sum of all Y components
Finally, we can find the magnitude and direction of the net displacement using the following equations:
Magnitude = sqrt((Net X component)^2 + (Net Y component)^2)
Direction = atan2(Net Y component, Net X component)
This will give us the direction and distance for Bubba to return directly to the boat ramp.
To calculate the time it took Bubba to reach the spot where he realized he forgot his fishing pole, we can sum up the distances of each individual displacement and divide it by the speed:
Total distance = (121 + 104 + 211 + 135) meters
Time = Total distance / Speed
To calculate the time it takes for Bubba to return to the boat ramp on the most direct path, we can use the distance between the spot where he realized his mistake and the boat ramp:
Direct distance = sqrt((Net X component)^2 + (Net Y component)^2)
Time = Direct distance / Speed
By plugging in the values and performing the calculations, we can get the answers to these questions.