| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x_S = 22\,\text{m} – 5\,\text{m} = 17\,\text{m}\] | Sally is \(5\,\text{m}\) ahead of Mary, leaving her \(\Delta x_S = 17\,\text{m}\) to the finish. |
| 2 | \[\Delta x_S = v_{i,S} t + \tfrac{1}{2} a_S t^2\] | Kinematic displacement equation for Sally with constant acceleration. |
| 3 | \[17 = 5 t + \tfrac{1}{2}(-0.40)t^2\] | Substitute \(v_{i,S}=5\,\text{m/s}\) and \(a_S=-0.40\,\text{m/s}^2\). |
| 4 | \[0.20 t^2 – 5 t + 17 = 0\] | Rearrange into standard quadratic form \(at^2+bt+c=0\). |
| 5 | \[t = \frac{5 – \sqrt{25 – (4)(0.20)(17)}}{0.40} = 4.06\,\text{s}\] | Solve the quadratic; the smaller root gives the physical time to finish because speed remains positive. |
| 6 | \[\Delta x_M = v_{i,M} t + \tfrac{1}{2} a_M t^2\] | Kinematic displacement equation for Mary, who must finish in the same time \(t=4.06\,\text{s}\). |
| 7 | \[22 = 4(4.06) + \tfrac{1}{2} a_M (4.06)^2\] | Insert Mary’s initial speed \(v_{i,M}=4.0\,\text{m/s}\) and her distance \(\Delta x_M=22\,\text{m}\). |
| 8 | \[a_M = \frac{22 – 16.24}{0.5(16.48)} = 0.70\,\text{m/s}^2\] | Algebraically solve for Mary’s required constant acceleration. |
| 9 | \[\boxed{a_M \approx 0.70\,\text{m/s}^2}\] | Mary must accelerate at this rate to finish side-by-side with Sally. |
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A driver is driving at \( 40 \, \text{m/s} \) when the light turns red in front of her. It takes the driver \( 0.9 \, \text{s} \) to react and hit the brakes. After this, the car slows with an acceleration of \( 3.5 \, \text{m/s}^2 \). What is the total distance traveled by the car?
A baseball is tossed from street level by a student straight up at a speed of \(25.3 \text{ m/s}\). After reaching maximum height, it is caught by another student on the roof of a building, \(17.4 \text{ m}\) above the street. How long did this take?
An object is released from rest near the surface of a planet. The velocity of the object as a function of time is expressed in the following equation. \( v_y = (-3) \, \text{m/s}^2 \, t \) All frictional forces are considered to be negligible. What distance does the object fall \( 10 \) \( \text{s} \) after it is released from rest?
A stone is thrown vertically upwards with a speed of \( 20.0 \) \( \text{m/s} \). How fast is it moving when it reaches a height of \( 12.0 \) \( \text{m} \)?
A body moving in the positive \( x \) direction passes the origin at time \( t = 0 \). Between \( t = 0 \) and \( t = 1 \, \text{second} \), the body has a constant speed of \( 24 \, \text{m/s} \). At \( t = 1 \, \text{second} \), the body is given a constant acceleration of \( 6 \, \text{m/s}^2 \) in the negative \( x \) direction. The position \( x \) of the body at \( t = 11 \, \text{seconds} \) is
A boat is rowed directly upriver at a speed of \(2.5 \, \text{m/s}\) relative to the water. Viewers on the shore find that it is moving at only \(0.5 \, \text{m/s}\) relative to the shore. What is the speed of the river? Is it moving with or against the boat?
A 100-pound rock and a 1-pound metal arrow pointed downwards are dropped from height \( h \). Assuming there is no air resistance, which one hits the ground first and why?
An object travels along a path shown above, with changing velocity as indicated by vectors \( A \) and \( B \). Which vector best represents the net acceleration of the object from time \( t_A \) to \( t_B \)?
A ball is tossed directly upward. Its total time in the air is \( T \). Its maximum height is \( H \). What is its height after it has been in the air a time \( T/4 \)? Air resistance is negligible.
On Saturday, Ashley rode her bicycle to visit Maria. Maria’s house is directly east of Ashley’s. The graph shows how far Ashley was from her house after each minute of her trip.(Hint – Use the standard units of velocity (m/s) for all parts)
\(0.70\, \text{m/s}^2\)
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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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