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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[v_{\text{rel}} = v_{m} – v_{c} = 30 – 20 = 10\;\text{m/s}\] | The relative speed is the motorcycle’s speed minus the car’s speed. |
| 2 | \[v_{\text{rel}} > 0\] | A positive relative speed means the motorcycle is closing the gap. |
| 3 | \[\boxed{\text{Yes}}\] | Because the motorcycle is faster, it will eventually catch the car. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[t = \frac{\Delta x_{\text{initial}}}{v_{\text{rel}}} = \frac{30}{10} = 3\;\text{s}\] | Time required equals initial separation divided by relative speed. |
| 2 | \[\Delta x = v_{c}\, t = 20 \times 3 = 60\;\text{m}\] | Distance from the car’s start is its speed times the catch-up time. |
| 3 | \[\boxed{t = 3\;\text{s},\; \Delta x = 60\;\text{m}}\] | Final answers for part (b). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[x_{c}(2) = 20 \times 2 = 40\;\text{m}\] | Car’s position after the first \(2\) seconds. |
| 2 | \[x_{m}(2) = 30 \times 2 – 30 = 30\;\text{m}\] | Motorcycle’s position after \(2\) seconds, relative to the car’s starting position. |
| 3 | \[\Delta x_{0} = x_{m} – x_{c} = -10\;\text{m}\] | So the motorcycle is still \(10 ~\text{m}\) behind when acceleration starts (at the \(2\) second mark). So let’s make an equation, for each vehicle, to see if and where they will meet. |
| 4 | \[x_{c} = 40 + 20 t_{1} + \tfrac{1}{2}(1) t_{1}^{2}\] | Car’s position for time \(t_{1}\) seconds after the \(2\) second mark. The equation shows that the car accelerates from the \(40\) meter mark. |
| 5 | \[x_{m} = 30 + 30 t_{1}\] | Motorcycle’s position for time \(t_{1}\) seconds after the \(2\) second mark. The equation shows the motorcycle continues at constant speed from its \(30\) meter mark. |
| 6 | \[x_{m}=x_{c}\;\Rightarrow\;-10 + 10 t_{1} – 0.5 t_{1}^{2}=0\] | We want to find when and if the vehicles meet. So set position equations, from step 4 and 5, equal to each other to find the catch-up (meet-up) time. |
| 7 | \[t_{1}^{2}-20 t_{1}+20 = 0\] | Multiply both sides by \(-2\) to simplify and rearrange to a quadratic equation. |
| 8 | \[t_{1}=\frac{20 \pm \sqrt{400-80}}{2}=1.06\;\text{s}\;\text{(smaller root)}\] | The larger root occurs after the car overtakes; only the smaller root is physical. |
| 9 | \[t = 2 + t_{1} = 2 + 1.06 = 3.06\;\text{s}\] | Total time from the initial start. |
| 10 | \[\Delta x = 40 + 20(1.06) + 0.5(1.06)^{2} \approx 61.7\;\text{m}\] | Distance from the car’s starting point where they meet. |
| 11 | \[\boxed{\text{Yes},\; t = 3.06\;\text{s},\; \Delta x = 61.7\;\text{m}}\] | The motorcycle still catches the car despite the acceleration. |
Just ask: "Help me solve this problem."
Which of the following graphs represent an object at rest?
Why is the stopping distance of a truck much shorter than for a train going the same speed? Hint: try deriving a formula or stopping distance.
A \(10 \, \text{kg}\) box is pushed to the right by an unknown force at an angle of \(25^\circ\) below the horizontal while a friction force of \(50 \, \text{N}\) acts on the box as well. The box accelerates from rest and travels a distance of \(4 \, \text{m}\) where it is moving at \(3 \, \text{m/s}\).
A baseball is seen to pass upward by a window with a vertical speed of \( 14 \) \( \text{m/s} \). If the ball was thrown by a person \( 18 \) \( \text{m} \) below on the street, determine the following.
A blue ball is thrown upward with a velocity of \( 9 \) \( \text{m/s} \) upward from the top of a high cliff. At the same time, a red ball is dropped from the same spot. The red ball is observed to hit the ground below exactly \( 1 \) \( \text{s} \) before the blue ball. How high is the cliff?
\( \text{Yes} \)
\( t = 3\,\text{s},\; \Delta x = 60\,\text{m} \)
\( \text{Yes},\; t = 3.06\,\text{s},\; \Delta x = 61.7\,\text{m} \text{from the car’s start}\)
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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] |
General Metric Conversion Chart
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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