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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."
Person A throws a ball horizontally from a cliff \( 20 \) \( \text{m} \) tall at \( 12 \) \( \text{m/s} \). Person B is running to the right on the ground and catches the ball at the same height it would’ve landed after running \( 15 \) \( \text{m} \). How fast was Person B running?
There are two cables that lift an elevator, each with a force of \(10{,}000 \, \text{N}\). The \(1{,}000 \, \text{kg}\) elevator is lifted from the first floor and accelerates over \(10 \, \text{m}\) until it reaches its top speed of \(6 \, \text{m/s}\). What is the mass of the people in the elevator?
A driver is traveling at a speed of \( 18.0 \) \( \text{m/s} \) when she sees a red light ahead. Her car is capable of decelerating at a rate of \( 3.65 \) \( \text{m/s}^2 \). If it takes her \( 0.350 \) \( \text{s} \) to get the brakes on and she is \( 20.0 \) \( \text{m} \) from the intersection when she sees the light, will she be able to stop in time? How far from the beginning of the intersection will she be, and in what direction?
You throw a ball straight upward. It leaves your hand at \( 20 \) \( \text{m/s} \) and slows at a steady rate until it stops at the peak. The ball then comes back down, speeding up steadily until it hits the ground with the same speed it left your hand. Draw the velocity vs. time graph or explain it in terms of functions.
\( \text{Yes} \)
\( t = 3\,\text{s},\; \Delta x = 60\,\text{m} \)
\( \text{Yes},\; t = 3.06\,\text{s},\; \Delta x = 61.7\,\text{m} \text{from the car’s start}\)
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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