A shopper pushes a cart 43 m south down one aisle and then turns 90.0° west and moves 15 m. He then makes another 90.0° turn and moves 10 m.
(a) What is the largest possible displacement of the shopper?
______ m _____° counterclockwise from west
(b) What is the smallest possible displacement of the shopper?
______ m _____° counterclockwise from west
start the shopper at (0,0)
After pushing the cart 43 m south, he's now at (0, -43)
After turning west and pushing the cart 15 m, he's now at (-15, -43)
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 (-15, -53)
north will result in the smallest displacement from the origin . . . then he'd be at (-15, -33)
The distance from the origin for the largest displacement is (15^2 + 53^2)^0.5
Let the angle counterclockwise from west be x: then
tan x = 53/15
The distance from the origin for the smallest displacement is (15^2 + 33^2)^0.5
at an angle:
tan x = 33/15
To find the largest and smallest displacement of the shopper, we can use vector addition. Let's break down the shopper's movements step-by-step.
(a) Largest possible displacement:
1. Moving 43 m south: This can be represented as a vector with a magnitude of 43 m, pointing in the south direction.
- Magnitude: 43 m
- Direction: South (+90° counterclockwise from west)
2. Turning 90.0° west and moving 15 m: This can be represented as a vector with a magnitude of 15 m, pointing in the west direction.
- Magnitude: 15 m
- Direction: West (0° counterclockwise from west)
3. Turning 90.0° and moving 10 m: This can be represented as a vector with a magnitude of 10 m, pointing in the north direction.
- Magnitude: 10 m
- Direction: North (-90° counterclockwise from west)
To find the largest displacement, we need to add these vectors together. Adding the vectors, we get:
Resultant magnitude: √((43 m)^2 + (15 m)^2 + (10 m)^2) = √(1849 m^2 + 225 m^2 + 100 m^2) = √(2174 m^2) ≈ 46.64 m
Resultant direction (counterclockwise from west): tan^(-1)((-10 m + 15 m) / -43 m) = tan^(-1)(5 m / -43 m) ≈ -6.31°
Therefore, the largest possible displacement of the shopper is approximately 46.64 m counterclockwise from west.
(b) Smallest possible displacement:
To find the smallest displacement, we need to subtract the vectors in the opposite order (taking negative of the direction of the vectors).
1. Moving 43 m south: Same as before.
- Magnitude: 43 m
- Direction: South (+90° counterclockwise from west)
2. Turning 90.0° west and moving 15 m: Same as before, but opposite direction.
- Magnitude: -15 m
- Direction: East (180° counterclockwise from west)
3. Turning 90.0° and moving 10 m: Same as before, but opposite direction.
- Magnitude: -10 m
- Direction: South (+90° counterclockwise from west)
Adding these vectors together, we get:
Resultant magnitude: √((43 m)^2 + (-15 m)^2 + (-10 m)^2) = √(1849 m^2 + 225 m^2 + 100 m^2) = √(2174 m^2) ≈ 46.64 m
Resultant direction (counterclockwise from west): tan^(-1)((-10 m - 15 m) / -43 m) = tan^(-1)(-25 m / -43 m) ≈ 30.10°
Therefore, the smallest possible displacement of the shopper is approximately 46.64 m counterclockwise from west.
To find the largest and smallest possible displacements of the shopper, we need to analyze the given directions and distances.
First, let's represent the directions using a coordinate system. We'll assume the starting point is the origin (0,0).
(a) Largest possible displacement:
The shopper moves 43 m south, which means a negative 43 in the y-direction (-43,0).
Then the shopper turns 90.0° west and moves 15 m. We can represent this as (-43,-15).
Finally, the shopper makes another 90.0° turn and moves 10 m, resulting in a displacement of (-43,-15-10) = (-43,-25).
To find the magnitude and direction of this displacement, we can use the Pythagorean theorem and trigonometry:
Magnitude = √((-43)^2 + (-25)^2) ≈ 50.6 m
Direction = arctan((-25)/(-43)) ≈ 30.1°
Therefore, the largest possible displacement of the shopper is approximately 50.6 m counterclockwise from west.
(b) Smallest possible displacement:
The shopper moves 43 m south, resulting in a displacement of (0,-43).
Then the shopper turns 90.0° west and moves 15 m, which can be represented as (-15,-43).
Finally, the shopper makes another 90.0° turn and moves 10 m, resulting in a displacement of (-15,-43-10) = (-15,-53).
To find the magnitude and direction of this displacement:
Magnitude = √((-15)^2 + (-53)^2) ≈ 55.9 m
Direction = arctan((-53)/(-15)) ≈ 74.2°
Therefore, the smallest possible displacement of the shopper is approximately 55.9 m counterclockwise from west.