| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[\text{Downward positive: } a=+g,\; v_i=+23\] | Define downward as the positive direction so that both \(a = g\) and \(v_i\) are positive. |
| 2 | \[\Delta x = v_i t + \frac{1}{2} g t^2\] | Use the constant-acceleration kinematic relation for displacement. |
| 3 | \[200 = 23 t + 4.9 t^2\] | Insert \(\Delta x = 200\,\text{m},\; v_i = 23\,\text{m/s},\; g = 9.8\,\text{m/s}^2\). |
| 4 | \[4.9 t^2 + 23 t – 200 = 0\] | Arrange terms to standard quadratic form \(at^2+bt+c=0\). |
| 5 | \[t = \frac{-23 + \sqrt{23^2 + 4(4.9)(200)}}{2(4.9)}\] | Apply the quadratic formula and keep the positive root (object cannot have negative time). |
| 6 | \[\sqrt{23^2 + 4(4.9)(200)} = \sqrt{4449} \approx 66.7\] | Calculate the discriminant value. |
| 7 | \[t = \frac{-23 + 66.7}{9.8} \approx 4.46\] | Compute \(t\) from the evaluated numbers. |
| 8 | \[\boxed{t \approx 4.46\,\text{s}}\] | State the time required to reach the ground. |
| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[v_x^2 = v_i^2 + 2 g \Delta x\] | Relate final and initial speeds using constant-acceleration kinematics. |
| 2 | \[v_x^2 = 23^2 + 2(9.8)(200) = 4449\] | Substitute \(v_i = 23\,\text{m/s},\; g = 9.8\,\text{m/s}^2,\; \Delta x = 200\,\text{m}\). |
| 3 | \[v_x = \sqrt{4449} \approx 66.7\] | Take the square root; sign is positive because downward is defined positive. |
| 4 | \[\boxed{v_x \approx 66.7\,\text{m/s (down)}}\] | Report the magnitude and indicate downward direction. |
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The first \(10 \, \text{meters}\) of a \(100 \, \text{meter}\) dash are covered in \(2 \, \text{seconds}\) by a sprinter who starts from rest and accelerates with a constant acceleration. The remaining \(90 \, \text{meters}\) are run with the same velocity the sprinter had after \(2 \, \text{seconds}\).
A rocket, initially at rest, is fired vertically upward with an acceleration of \( 12.0 \, \text{m/s}^2 \). At an altitude of \( 1.00 \, \text{km} \), the rocket engine cuts off. Drag is negligible.
Can an object have \( 0 \) velocity and nonzero acceleration at the same time? Give two examples.
A mine shaft is known to be 57.8 m deep. If you dropped a rock down the shaft, how long would it take for you to hear the sound of the rock hitting the bottom of the shaft knowing that sound travels at a constant velocity of 345 m/s?
A mass moving with a constant speed \( u \) encounters a rough surface and comes to a stop. The mass takes a time \( t \) to stop after encountering the rough surface. The coefficient of dynamic friction between the rough surface and the mass is \( 0.40 \). Which of the following expressions gives the initial speed \( u \)?
A ball is dropped off a high cliff, and \( 2 \) \( \text{s} \) later another ball is thrown vertically downward with an initial speed of \( 30 \) \( \text{m/s} \). How long will it take the second ball to overtake the first?
A gun can fire a bullet to height \( h \) when fired straight up. If the same gun is pointed at an angle of \( 45^\circ \) from the vertical, what is the new maximum height of the projectile?
A ball is dropped from a window \(10 \, \) above the sidewalk. Determine the time it takes for the ball to fall to the sidewalk.
At time \( t = 0 \), a cart is at \( x = 10 \, \text{m} \) and has a velocity of \( 3 \, \text{m/s} \) in the \( -x \) direction. The cart has a constant acceleration in the \( +x \) direction with magnitude \( 3 \, \text{m/s}^2 < a < 6 \, \text{m/s}^2 \). Which of the following gives the possible range of the position of the cart at \( t = 1 \, \text{s} \)?
\(4.46\,\text{s}\)
\(66.7\,\text{m/s (downward)}\)
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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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