| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( y = y_0 + v_0 t + \frac{1}{2} a t^2 \) | Use the kinematic equation for vertical motion to find the time. Here, \( y \) is the final position, \( y_0 \) is the initial position, \( v_0 \) is the initial velocity, \( a \) is the acceleration due to gravity, and \( t \) is the time. |
| 2 | \( y = 0 \, \text{m}\), \( y_0 = 10 \, \text{m} \), \( v_0 = 0 \, \text{m/s} \), \( a = -9.8 \, \text{m/s}^2 \) | Define the values for each variable. The final position \( y \) is 0 m (sidewalk level), initial height \( y_0 \) is 10 m, initial velocity \( v_0 \) is 0 m/s (dropped from rest), and acceleration \( a \) is \( -9.8 \, \text{m/s}^2 \) (downwards). |
| 3 | \( 0 = 10 + 0 \cdot t + \frac{1}{2} \cdot (-9.8) \cdot t^2 \) | Plug in the known values into the kinematic equation. |
| 4 | \( 0 = 10 – 4.9 t^2 \) | Simplify the equation by eliminating the term with zero initial velocity. |
| 5 | \( 4.9 t^2 = 10 \) | Rearrange the equation to isolate the term with \( t^2 \). |
| 6 | \( t^2 = \frac{10}{4.9} \) | Divide both sides of the equation by 4.9 to solve for \( t^2 \). |
| 7 | \( t^2 \approx 2.04 \) | Calculate the result of the division. |
| 8 | \( t \approx \sqrt{2.04} \) | Take the square root of both sides to solve for \( t \). |
| 9 | \( t \approx 1.43 \, \text{s} \) | Calculate the square root to find the time. |
| 10 | \(\mathbf{t \approx 1.43 \, \text{s}}\) | Final answer. |
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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?
An object of unknown mass is acted upon by multiple forces:
The coefficients of friction are \(\mu_s = 0.6\) and \(\mu_k = 0.2\). Starting from rest, the object travels \(10 \, \text{m}\) in \(4.5 \, \text{s}\). What is the mass of the unknown object?
Which pair of quantities will always have the same magnitude if motion is in a straight line and in one direction?
A coin is dropped from a hot air-balloon that is \(250 \, \text{m}\) above the ground rising at \(11 \, \text{m/s}\) upwards. For the coin, assume up is positive and find the following:
A teacher walks the following path in \( 10 \) \( \text{s} \): \( 2 \) \( \text{m} \) south, \( 4 \) \( \text{m} \) east, \( 2 \) \( \text{m} \) north, \( 4 \) \( \text{m} \) west. What is the teacher’s average velocity?
A helicopter is ascending vertically with a speed of \( 5.40 \) \( \text{m/s} \). At a height of \( 105 \) \( \text{m} \) above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground?
An object can move upward while having a downward acceleration.
On another planet, a ball is in free fall after being released from rest at time \( t = 0 \). A graph of the height of the ball above the planet’s surface as a function of time \( t \) is shown. The acceleration due to gravity on the planet is most nearly
Mary and Sally are in a foot race. When Mary is \( 22 \) \( \text{m} \) from the finish line, she has a speed of \( 4.0 \) \( \text{m/s} \) and is \( 5.0 \) \( \text{m} \) behind Sally, who has a speed of \( 5.0 \) \( \text{m/s} \). Sally thinks she has an easy win and, during the remaining portion of the race, decelerates at a constant rate of \( 0.40 \) \( \text{m/s}^2 \) until she reaches the finish line. What constant acceleration must Mary maintain during the remaining portion of the race if she wishes to cross the finish line side-by-side with Sally?
Toy car W travels across a horizontal surface with an acceleration of \( a_w \) after starting from rest. Toy car Z travels across the same surface toward car W with an acceleration of \( a_z \), after starting from rest. Car W is separated from car Z by a distance \( d \). Which of the following pairs of equations could be used to determine the location on the horizontal surface where the two cars will meet, and why?
1.4 seconds
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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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