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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | Choose east as +x. Set the red car’s start at the origin. \[x_R(0)=0,\quad v_{R0}=0,\quad a_R=+3.5,\text{m/s}^2\] Blue car starts 600 m to the east: \[x_B(0)=600,\text{m},\quad v_{B0}=-15,\text{m/s},\quad a_B=-1.2,\text{m/s}^2\] |
Since the blue car moves west, its velocity is negative. Interpreting “accelerating at 1.2 m/s²” while traveling west means its acceleration is also westward (negative) in this coordinate system. |
| 2 | Constant-acceleration position functions: \[x_R(t)=x_R(0)+v_{R0}t+\tfrac{1}{2} a_R t^2=\tfrac{1}{2}(3.5)t^2=1.75t^2\] \[x_B(t)=x_B(0)+v_{B0}t+\tfrac{1}{2} a_B t^2=600-15t-0.6t^2\] |
Use \(x=x_0+v_0 t+\tfrac{1}{2} a t^2\) for each car. |
| 3 | They meet when \(x_R(t)=x_B(t)\): \[1.75t^2=600-15t-0.6t^2\] \[2.35t^2+15t-600=0\] |
Meeting means same position at the same time; rearranging gives a quadratic in \(t\). |
| 4 | Quadratic formula for \(at^2+bt+c=0\) with \(a=2.35\), \(b=15\), \(c=-600\): \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] Discriminant: \[D=b^2-4ac=15^2-4(2.35)(-600)=225+5640=5865\] |
Solve the quadratic exactly; keep sufficient precision to avoid inconsistent positions later. |
| 5 | Compute \(\sqrt{5865}\) accurately: \[\sqrt{5865}\approx 76.583\] Then \[t=\frac{-15+76.583}{2(2.35)}=\frac{61.583}{4.7}\approx 13.103,\text{s}\] |
Choose the (+) root because time must be positive; the other root is negative and not physical. |
| 6 | Meeting position (use either car; results should match when using the same \(t\)): \[x= x_R(t)=1.75t^2\approx 1.75(13.103)^2\] \[(13.103)^2\approx 171.683\quad\Rightarrow\quad x\approx 1.75(171.683)\approx 300.45,\text{m}\] Check with blue car: \[x_B=600-15t-0.6t^2\approx 600-15(13.103)-0.6(171.683)\approx 300.45,\text{m}\] |
Using consistent precision eliminates the earlier mismatch; both equations must give the same meeting point. |
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An object is moving in the \( +x \) direction and begins to slow down. What must be true about its acceleration?
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?
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 spacecraft accelerates at a rate of \(20.0 \, \text{m/s}^2\).
A \( 1000 \) \( \text{kg} \) car is traveling east at \( 20 \) \( \text{m/s} \) when it collides perfectly inelastically with a northbound \( 2000 \) \( \text{kg} \) car traveling at \( 15 \) \( \text{m/s} \). If the coefficient of kinetic friction is \( 0.9 \), how far, and at what angle do the two cars skid before coming to a stop?
An airplane is traveling \( 900. \) \( \text{km/h} \) in a direction \( 38.5^\circ \) west of north.
A car is driving to the right at \( 20 \) \( \text{m/s} \). A motorcycle starts \( 30 \) \( \text{m} \) behind the car and is moving at \( 30 \) \( \text{m/s} \) in the same direction.
Runner A begins a \( 100 \)-meter race at time \( t = 0 \) and runs at a constant speed of \( 6.0 \) \( \text{m/s} \). Runner B starts the same race \( 3 \) seconds later but runs at \( 9.0 \) \( \text{m/s} \).
\(t\approx 13.10\,\text{s}\), \(x\approx 300.45\,\text{m}\) east of the red car’s starting point (i.e., \(600-300.45\approx 299.55\,\text{m}\) west of the blue car’s starting point).
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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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