| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_x = v_i + at_1\) | First, use the equation of motion to express final velocity \(v_x\) at \(t_1\). Since the sled starts from rest, \(v_i = 0\), so \(v_x = at_1\). |
| 2 | \(v_x = 13.0 \, \text{m/s}^2 \times t_1\) | Substitute the given acceleration \(a = 13.0 \, \text{m/s}^2\). |
| 3 | \(d_1 = \frac{1}{2} a t_1^2\) | Use the equation for displacement under constant acceleration to find \(d_1\), the distance covered during the acceleration phase. |
| 4 | \(d_1 = \frac{1}{2} \times 13.0 \, \text{m/s}^2 \times t_1^2\) | Substitute the value of \(a\). |
| 5 | \(d_2 = v_x(t_2 – t_1)\) | Use the equation for displacement under constant velocity for the distance traveled from \(t_1\) to \(t_2\). |
| 6 | \(d_2 = \left(13.0 \, \text{m/s}^2 \times t_1\right)\left(90.0\, \text{s} – t_1\right)\) | Substitute \(v_x\) and the time intervals into the equation for the second part of the journey. |
| 7 | \(d_1 + d_2 = 5300 \, \text{m}\) | Use the total distance formula, combining both displacements to equal the total given distance. |
| 8 | \(\frac{1}{2} \times 13.0 \, \text{m/s}^2 \times t_1^2 + (13.0 \, \text{m/s}^2 \times t_1)(90.0 \, \text{s} – t_1) = 5300 \, \text{m}\) | Combine the equations for \(d_1\) and \(d_2\). |
| 9 | \(6.5 t_1^2 + 1170 t_1 – 13 t_1^2 = 5300\) | Simplify the equation by combining like terms. |
| 10 | \(-6.5 t_1^2 + 1170 t_1 – 5300 = 0\) | Combine like terms and solve the quadratic equation via the quadratic formula or graphing. |
| 11 | \(t_1 = 4.65 \, \text{s}\) | Final answer. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(t_2 + t_1 = 90.0 \, \text{s}\) | The total time of travel is the time of the \(t_1\) and \(t_2\) added together. |
| 2 | \(t_2 = 90.0 \, \text{s} – t_1 \, \text{s}\) | Rearrange and solve for \(t_2\). |
| 3 | \(t_2 = 90.0 \, \text{s} – 4.58 \, \text{s}\) | Substitute the calculated value of \(t_1 \) found in part a. |
| 4 | \(t_2 = 85.35 \, \text{s}\) | Final \(t_2\) value. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v = a t_1\) | Use the equation for final velocity under constant acceleration. This comes from the kinematic formula \(v = v_0 + a t_1\), where \(v_0\) is zero. |
| 2 | \(v = 13.0 \, \text{m/s}^2 \times 4.65 \, \text{s}\) | Substitute the value of \(a\) and \(t_1\) into the equation. |
| 3 | \(v = 60.5 \, \text{m/s}\) | Calculate the velocity. |
– (a) \( \boxed{t_1 = 4.58 \, \text{s}} \)
– (b) \( \boxed{t_2 = 85.42\, \text{s}} \)
– (c) \( \boxed{v = 59.54 \, \text{m/s}} \)
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A ball is launched horizontally from a height. At the same time, another ball is dropped vertically from the same height. Which hits the ground first?
A student walks \( 3 \) \( \text{m} \) east, then \( 4 \) \( \text{m} \) west in \( 7 \) \( \text{s} \). What is their displacement and average velocity?
A runner completes one full lap around a \( 400 \) \( \text{m} \) track in \( 100 \) \( \text{s} \). What is their average velocity?
A car’s velocity increases as follows each second: \( 2 \) \( \text{m/s} \), \( 4 \) \( \text{m/s} \), \( 6 \) \( \text{m/s} \), \( 8 \) \( \text{m/s} \). This pattern shows that the car is:
A \(30 \, \text{g}\) bullet is fired with a speed of \(500 \, \text{m/s}\) into a wall.
A car increases its forward velocity uniformly from \(40 ~ \text{m/s}\) to \(80 ~ \text{m/s}\) while traveling a distance of \(200 ~ \text{m}\). What is its acceleration during this time?
A body starting from rest moves along a straight line under the action of a constant force. After traveling a distance \( d \) the speed of the body is \( v \). The speed of the body when it has travelled a distance \( \dfrac{d}{2} \) from its initial position is
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 \( t_B \)?
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 \(\text{m/s}\) for all parts)
\( t_1 = 4.65 \, \text{s} \)
\( t_2 = 85.35 \, \text{s} \)
\( v = 60.5 \, \text{m/s} \)
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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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