| Derivation/Formula | Reasoning |
|---|---|
| $$\frac{1}{2} m v_i^2 = \frac{1}{2} m v_x^2 + m g \Delta x$$ | Mechanical energy is conserved; initial kinetic changes into kinetic plus gravitational potential after rising by \(\Delta x = 13\text{ m}\). |
| $$v_x^2 = v_i^2 – 2 g \Delta x$$ | Algebraically solve the previous equation for \(v_x\). |
| $$v_x = \sqrt{v_i^2 – 2 g \Delta x}$$ | Take the square root to obtain speed magnitude. |
| $$v_x = \sqrt{(24.0)^2 – 2 (9.8)(13.0)}\;\text{m/s}$$ | Insert \(v_i = 24.0\text{ m/s}\), \(g = 9.8\text{ m/s}^2\), and \(\Delta x = 13.0\text{ m}\). |
| $$\boxed{v_x \approx 17.9\ \text{m/s}}$$ | Numeric result; direction can be up or down, magnitude shown. |
| Derivation/Formula | Reasoning |
|---|---|
| $$\Delta x = v_i t + \frac{1}{2} a t^2$$ | Kinematic displacement equation with upward positive. |
| $$13 = 24 t – 4.9 t^2$$ | Substitute \(\Delta x = 13\), \(v_i = 24\text{ m/s}\), \(a = -g = -9.8\text{ m/s}^2\). |
| $$4.9 t^2 – 24 t + 13 = 0$$ | Rearrange into standard quadratic form. |
| $$t = \frac{24 \pm \sqrt{24^2 – 4(4.9)(13)}}{2(4.9)}$$ | Apply the quadratic formula to solve for \(t\). |
| $$t = \frac{24 \pm 17.9}{9.8}$$ | Evaluate the discriminant \(\sqrt{321.2} \approx 17.9\). |
| $$t_1 \approx 0.62\ \text{s},\qquad t_2 \approx 4.28\ \text{s}$$ | Compute the two positive roots. |
| $$\boxed{t = 0.62\ \text{s}\ \text{or}\ 4.28\ \text{s}}$$ | Both times are physically valid. |
| Derivation/Formula | Reasoning |
|---|---|
| $$t_1 \text{ (ascending)},\; t_2 \text{ (descending)}$$ | The quadratic gives two roots because the stone passes the \(13\text{ m}\) level once while going up and again while coming down after reaching its peak. |
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A car moving at 30 m/s makes a head-on collision with a stone wall. From what height would the car have to fall in order to make an equally hard collision with the ground?
The displacement \( x \) of an object moving in one dimension is shown above as a function of time \( t \). The acceleration of this object must be
A rocket is sent to shoot down an invading spacecraft that is hovering at an altitude of \( 1500 \, \text{m} \). The rocket is launched with an initial velocity of \( 180 \, \text{m/s} \). Find the following:
A rubber ball bounces on the ground. After each bounce, the ball reaches one-half the height of the bounce before it. If the time the ball was in the air between the first and second bounce was 1 second. What would be the time between the second and third bounce?
A projectile of mass 0.750 kg is shot straight up with an initial speed of 18.0 m/s.
An airplane accelerates down a runway at \( 10 \, \text{m/s}^2 \). It reaches a final velocity of \( 200 \, \text{m/s} \) until it finally lifts off the ground. Determine the distance traveled before takeoff.
In which of the following is the particle’s acceleration constant?
An object undergoes constant acceleration. Starting from rest, the object travels \( 5 \, \text{m} \) in the first second. Then it travels \( 15 \, \text{m} \) in the next second. What total distance will be covered after the 3rd second?
Two objects are dropped from rest from the same height. Object \( A \) falls through a distance \( d_A \) during a time \( t \), and object \( B \) falls through a distance \( d_B \) during a time \( 2t \). If air resistance is negligible, what is the relationship between \( d_A \) and \( d_B \)?
A ball is thrown straight up. What are the velocity and acceleration of the ball at the highest point in its path?
a: \(17.9\ \text{m/s}\)
b: \(0.62\ \text{s}\) or \(4.28\ \text{s}\)
c: \(\text{Upward and downward pass}\)
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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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