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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] v_f^2 = v_i^2 + 2ad [/katex] | Use the kinematic equation relating final velocity [katex] v_f [/katex], initial velocity [katex] v_i [/katex], acceleration [katex] a [/katex], and distance [katex] d [/katex]. |
| 2 | [katex] v_f = 0 [/katex] | Since the car skids to a stop, the final velocity [katex] v_f [/katex] is 0. |
| 3 | [katex] 0 = v_i^2 + 2a d [/katex] | Substitute [katex] v_f = 0 [/katex] into the kinematic equation. |
| 4 | [katex] d = \frac{-v_i^2}{2a} [/katex] | Solve the equation for distance [katex] d [/katex]. |
| 5 | [katex] d_1 = \frac{-v_i^2}{2a} [/katex] | Express the initial distance [katex] d_1 [/katex] using the initial speed [katex] v_i [/katex]. |
| 6 | [katex] v_f = 0, v_i = 2v_i \rightarrow (2v_i)^2 [/katex] | Consider the scenario where the initial speed is twice as fast, i.e.,[katex] v_i = 2v_i [/katex]. Substituting this in the same equation. |
| 7 | [katex] d_2 = \frac{-(2v_i)^2}{2a} [/katex] | Solve for the new distance [katex] d_2 [/katex] using the new initial speed [katex] 2v_i [/katex]. |
| 8 | [katex] d_2 = \frac{-4v_i^2}{2a} = 4 \left( \frac{-v_i^2}{2a} \right) = 4d_1 [/katex] | By simplifying the resulting equation, we see that the new stopping distance is 4 times the initial stopping distance. |
| 9 | 4 times as far: (a) | Therefore, if the car was traveling twice as fast, it would skid 4 times as far. |
| 10 | (b) The same distance | Incorrect: This contradicts the earlier derivation that shows the distance scales quadratically with speed. |
| 11 | (c) 2 times as far | Incorrect: This would be true if distance scaled linearly with speed, which it does not. |
| 12 | (d) The mass of the car must be known | Incorrect: The stopping distance in the given context is independent of the mass of the car. |
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A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s. Then it slows down at a steady rate of 2.0 m/s2 for 4.0 s. What is the final speed of the car?
A rock is dropped from the top of a tall tower. Half a second later another rock, twice as massive as the first, is dropped. Ignoring air resistance and using ONLY simple kinematics (DO NOT use energy to explain this). Explain it like you would to a 5th grader and select the correct choice:
Two students start \( 100 \) \( \text{m} \) apart.
• Student A walks to the right at \( 2 \) \( \text{m/s} \).
• Student B walks to the left at \( 3 \) \( \text{m/s} \).
At what time do the students meet, and how far has each student walked when they collide?
An object is released from rest near the surface of a planet. The velocity of the object as a function of time is expressed in the following equation. \( v_y = (-3) \, \text{m/s}^2 \, t \) All frictional forces are considered to be negligible. What distance does the object fall \( 10 \) \( \text{s} \) after it is released from rest?
A car decelerates from \( 25 \, \text{m/s} \) to \( 5 \, \text{m/s} \) at \( 10 \, \text{m/s}^2 \). How far does the car travel during this deceleration?
A car travels east at a steady \( 30 \) \( \text{m/s} \) for \( 5 \) \( \text{s} \). What is its acceleration during this motion?
A ball is dropped from a window [katex]10 \, [/katex] above the sidewalk. Determine the time it takes for the ball to fall to the sidewalk.
A spacecraft accelerates at a rate of \(20.0 \, \text{m/s}^2\).
A driver is driving at \( 40 \, \text{m/s} \) when the light turns red in front of her. It takes the driver \( 0.9 \, \text{s} \) to react and hit the brakes. After this, the car slows with an acceleration of \( 3.5 \, \text{m/s}^2 \). What is the total distance traveled by the car?
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\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}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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