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A car accelerates uniformly from rest to 29.4 m/s in 6.93 s along a level stretch of road. Ignoring friction, determine the average power (in both watts and horse power) required to accelerate the car if

- The weight of the car is 6.35 x 10
^{3}N - The weight of the car is 1.11 × 10
^{4}N

*Assume that 1 horsepower is 745.7 Watts. *

- 40,367 Watts; 54 HP
- 70,564 Watts; 95 HP

Explanation

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**Common Values for Both Cases:**

- Final velocity (v): 29.4 m/s
- Time (t): 6.93 s
- Conversion factor: 1 HP = 745.7 Watts

**Case 1: Weight = 6.35 × 10³ N**

Step | Formula Derivation | Reasoning |
---|---|---|

1 | \Delta KE = \frac{1}{2}mv^2 – \frac{1}{2}mu^2 | Work is equal to the change in kinetic energy (\Delta KE) calculation, where m is mass, v is final velocity, and u is initial velocity (zero). |

2 | m = \frac{\text{Weight}}{g} | Calculating mass from weight. |

3 | P = \frac{\Delta KE}{t} | Average power (P) is change in kinetic energy over time. |

4 | P_{\text{HP}} = \frac{P}{745.7} | Convert power to horsepower. |

You can repeat the same steps for case 2.

Let’s perform calculation for both cases.

**Case 1: Weight = 6.35 × 10³ N**

- Average Power: \boxed{40367.90, \text{Watts}}
- In Horsepower: \boxed{54, \text{HP}}

**Case 2: Weight = 1.11 × 10⁴ N**

- Average Power: \boxed{70564.36, \text{Watts}}
- In Horsepower: \boxed{95, \text{HP}}

- Statistics

Find the escape speed from a planet of mass 6.89 x 10^{25} kg and radius 6.2 x 10^{6} m.

- 40,367 Watts; 54 HP
- 70,564 Watts; 95 HP

**A car accelerates uniformly from rest to 29.4 m/s in 6.93 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate…**

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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 .
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