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
Assume that 1 horsepower is 745.7 Watts.
Common Values for Both Cases:
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
Case 2: Weight = 1.11 × 10⁴ N
Home » 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
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 |