Menu

A seagull first flies 160 m North, the heads up 120.65 m 18.43° North of West. After you land,

- How far North and West have you moved
- What is the total distance of the seagulls path?
- What is the total displacement of the seagull

- 198 m North and 114.5 m West
- 280.65 m
- 228.6 m @ 60° North of West
*(note: you must give both magnitude and direction, as displacement is a vector)*

Explanation

- Statistics

A person is standing at the edge of the water and looking out at the ocean. The height of the person’s eyes above the water is *h* = 1.8 m, and the radius of the Earth is *R* = 6.38 × 10^{6} m. How far is it to the horizon (in meters)? In other words, find the distance d from the person’s eyes to the horizon.

*(Note: At the horizon the angle between the line of sight and the radius of the earth is 90°)*

Two racing boats set out from the same dock and speed away at the same constant speed of 101 km/h for half an hour (0.5 hr). Boat 1 is headed 27.6° south of west, and Boat 2 is headed 35.3° south of west, as shown in the graph above. During this half-hour calculate:

- How much farther west does Boat 1 travel, compared to Boat 2 in kilometers.
- How much farther south does Boat 2 travel, compared to the Boat 2 in kilometers.

Determine the sum of the 3 vectors given below. Give the resultant (R) in terms of (a) vector components (b) resultant vector.

Vectors:

\vec{A} = 26.5 m @ at 56° NW

\vec{B} = 44 m @ at 28° NE

\vec{C} = 31 m South

- 198 m North and 114.5 m West
- 280.65 m
- 228.6 m @ 60° North of West
*(note: you must give both magnitude and direction, as displacement is a vector)*

**A seagull first flies 160 m North, the heads up 120.65 m 18.43° North of West. After you land find…**

By continuing, you agree to the updated Terms of Sale, Terms of Use, and Privacy Policy.

What would you like us to add, improve, or change on this page? We listen to all your feedback!

- Scroll Horiztontally

Kinematics | Forces |
---|---|

\Delta x = v_i \cdot t + \frac{1}{2} a \cdot t^2 | F = m \cdot a |

v = v_i + a \cdot t | F_g = \frac{G \cdot m_1 \cdot m_2}{r^2} |

a = \frac{\Delta v}{\Delta t} | f = \mu \cdot N |

R = \frac{v_i^2 \cdot \sin(2\theta)}{g} |

Circular Motion | Energy |
---|---|

F_c = \frac{m \cdot v^2}{r} | KE = \frac{1}{2} m \cdot v^2 |

a_c = \frac{v^2}{r} | PE = m \cdot g \cdot h |

KE_i + PE_i = KE_f + PE_f |

Momentum | Torque and Rotations |
---|---|

p = m \cdot v | \tau = r \cdot F \cdot \sin(\theta) |

J = \Delta p | I = \sum m \cdot r^2 |

p_i = p_f | L = I \cdot \omega |

Simple Harmonic Motion |
---|

F = -k \cdot x |

T = 2\pi \sqrt{\frac{l}{g}} |

T = 2\pi \sqrt{\frac{m}{k}} |

Constant | Description |
---|---|

g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |

G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |

\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |

k | Spring constant, in \text{N/m} |

Variable | SI Unit |
---|---|

s (Displacement) | \text{meters (m)} |

v (Velocity) | \text{meters per second (m/s)} |

a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |

t (Time) | \text{seconds (s)} |

m (Mass) | \text{kilograms (kg)} |

Variable | Derived SI Unit |
---|---|

F (Force) | \text{newtons (N)} |

E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |

P (Power) | \text{watts (W)} |

p (Momentum) | \text{kilogram meters per second (kg·m/s)} |

\omega (Angular Velocity) | \text{radians per second (rad/s)} |

\tau (Torque) | \text{newton meters (N·m)} |

I (Moment of Inertia) | \text{kilogram meter squared (kg·m}^2\text{)} |

f (Frequency) | \text{hertz (Hz)} |

General Metric Conversion Chart

Conversion Example

Example of using unit analysis: Convert 5 kilometers to millimeters.

Start with the given measurement:

`\text{5 km}`

Use the conversion factors for kilometers to meters and meters to millimeters:

`\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}`

Perform the multiplication:

`\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}`

Simplify to get the final answer:

`\boxed{5 \times 10^6 \, \text{mm}}`

Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|

Pico- | p | 10^{-12} | 0.000000000001 |

Nano- | n | 10^{-9} | 0.000000001 |

Micro- | µ | 10^{-6} | 0.000001 |

Milli- | m | 10^{-3} | 0.001 |

Centi- | c | 10^{-2} | 0.01 |

Deci- | d | 10^{-1} | 0.1 |

(Base unit) | – | 10^{0} | 1 |

Deca- or Deka- | da | 10^{1} | 10 |

Hecto- | h | 10^{2} | 100 |

Kilo- | k | 10^{3} | 1,000 |

Mega- | M | 10^{6} | 1,000,000 |

Giga- | G | 10^{9} | 1,000,000,000 |

Tera- | T | 10^{12} | 1,000,000,000,000 |

- Some answers may be slightly off by 1% depending on rounding, etc.
- Answers will use different values of gravity. Some answers use 9.81 m/s
^{2}, and other 10 m/s^{2 }for calculations. - Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
- Bookmark questions that you can’t solve so you can come back to them later.
- Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!