| 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. |
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A train is moving to the right at \( 20 \) \( \text{m/s} \). A passenger on the train throws a ball horizontally to the left at \( 5 \) \( \text{m/s} \) (relative to the train).
You are standing on a bathroom scale in an elevator. The elevator starts from rest on the first floor and accelerates up to the third floor, \(12 \, \text{m}\) above, in a time of \(6 \, \text{s}\). The scale reads \(800 \, \text{N}\). What is the mass of the person?
An elevator of height \(h\) ascends with constant acceleration \(a\). When it crosses a platform, it has acquired a velocity \(u\). At this instant a bolt drops from the top of the elevator. Find the time for the bolt to hit the floor of the elevator. Give your answer in terms of \(h\), \(a\), and any constant.
Which graph below shows that one of the runners started 10 meters further ahead of the other? Assume the y-axis is measured in meters and the x-axis is measured in seconds.
At time \( t = 0 \) an object is traveling to the right along the \( +x \) axis at a speed of \( 10.0 \) \( \text{m/s} \) with acceleration \( -2.0 \) \( \text{m/s}^2 \). Which statement is true?
A person sitting in an enclosed train car, moving at constant velocity, throws a ball straight up into the air in her reference frame.
A car is traveling 20 m/s when the driver sees a child standing on the road. She takes 0.8 s to react then steps on the brakes and slows at 7.0 m/s2. How far does the car go before it stops?
Two identical metal balls are being held side by side at the top of a ramp. Alex lets one ball, \( A \), start rolling down the hill. A few seconds later, Alex’s partner, Bob, starts the second ball, \( B \), down the hill by giving it a push. Ball \( B \) rolls down an identical, parallel path to the first ball and passes it. At the instant ball \( B \) passes ball \( A \) (select all that apply):
An object is moving in a horizontal circle at a constant speed. Which of the following correctly describes the linear and angular velocities of the object between any point along the circular path?
\( \text{Yes} \)
\( t = 3\,\text{s},\; \Delta x = 60\,\text{m} \)
\( \text{Yes},\; t = 3.06\,\text{s},\; \Delta x = 61.7\,\text{m} \text{from the car’s start}\)
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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] |
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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