A shopper pushes a cart 41 m south down one aisle and then turns 90.0° west and moves 28 m. He then makes another 90.0° turn and moves 14 m.
(a) What is the largest possible displacement of the shopper?
____m ___° counterclockwise from west
start the shopper at (0,0)
After pushing the cart 41 m south, he's now at (0, -41)
After turning west and pushing the cart 28 m, he's now at (-28, -41)
Now he has to turn 90 degrees again. . . he will then be going either north or south. . . south will result in the largest displacement from the origin. . . now he's at (-28, -55)
The distance from the origin is (28^2 + 55^2)^0.5
Let the angle counterclockwise from west be x: then
tan x = 55/28
To find the largest possible displacement of the shopper, we can create a right triangle using the distances traveled in the north-south and east-west directions.
Let's break down the shopper's movements into their components:
- 41 m south
- 28 m west
- 14 m north
Now, let's calculate the displacement in the north-south (y) direction:
Total displacement in the y-direction = -41 m + 14 m = -27 m
Note that we use a negative sign since the displacement is in the opposite direction of the positive y-axis.
Next, let's calculate the displacement in the east-west (x) direction:
Total displacement in the x-direction = 28 m
Now, we can use the Pythagorean theorem to find the magnitude of the displacement:
Displacement (d) = √(x^2 + y^2)
d = √((28)^2 + (-27)^2)
d = √(784 + 729)
d ≈ √1513
d ≈ 38.90 m
So, the largest possible displacement of the shopper is approximately 38.90 meters.
To find the angle counterclockwise from the west, we can use trigonometry. Using the arctan function, we can find the angle:
angle = arctan(y/x)
angle = arctan(-27/28)
angle ≈ -45.65°
Thus, the largest possible displacement of the shopper is approximately 38.90 meters, counterclockwise from the west by about 45.65°.