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| 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) \( \boxed{v = 59.54 \, \text{m/s}} \)
Just ask: "Help me solve this problem."
The figure shows a graph of the position x of two cars, C and D, as a function of time t. According to this graph, which statements about these cars must be true? (There could be more
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Two students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down.
Which statements are not valid for a projectile? Take up as positive.
A 2,000 kg car collides with a stationary 1,000 kg car. Afterwards, they slide 6 m before coming to a stop. The coefficient of friction between the tires and the road is 0.7. Find the initial velocity of the 2,000 kg car before the collision?
Which graph below shows that one of the runners started 10 meters further ahead of the other? Assume the y-axis is measured in meters and the x-axis is measured in seconds.
\( t_1 = 4.65 \, \text{s} \)
\( t_2 = 85.35 \, \text{s} \)
\( v = 60.5 \, \text{m/s} \)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) | Â |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| Â | \(W = Fd \cos\theta\) |
| 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] |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.Â
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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