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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (a) Time and distance for the student to overtake the bus. | ||
| 1 | Define positions of student and bus:
– Student’s position: – Bus’s position: |
Established equations of motion for both the student and the bus. |
| 2 | Set positions equal to find overtaking time \( t \):
\( x_s(t) = x_b(t) \) |
Equated positions since they meet at the same point. |
| 3 | Plug in known values to form quadratic equation:
\( \dfrac{1}{2} (0.170) t^2 – 5.0 t + 40.0 = 0 \) |
Formed a quadratic equation in \( t \). |
| 4 | Solve the quadratic equation using the quadratic formula:
\( t = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a} \) |
Calculated the discriminant and solved for \( t \). |
| 5 | Compute the two possible times:
First solution: Second solution: |
Found two times when the student and bus meet. |
| 6 | Select the earlier time \( t = 9.553 \, \text{s} \):
Calculate the distance the student runs: |
Determined the time and distance for the student to overtake the bus. |
| (b) Speed of the bus when the student overtakes it. | ||
| 7 | Calculate bus’s velocity at \( t = 9.553 \, \text{s} \):
\( v_b = a_b t = 0.170 \times 9.553 \approx 1.624 \, \text{m/s} \) |
Found bus’s speed at the moment of overtaking. |
| (c) Sketch of \( x \) vs. \( t \) graph for both student and bus. | ||
| 8 | Description of the graph:
– Student’s path: Straight line starting from \( x = 0 \) with slope \( v_s = 5.0 \, \text{m/s} \). |
Visual representation of positions over time. |
| (d) Significance of the second time solution and bus’s speed at that point. | ||
| 9 | Second time from part (a): \( t = 49.27 \, \text{s} \):
– This is when the bus overtakes the student again. |
Explained the second intersection point and calculated bus’s speed. |
| (e) Will the student catch the bus at \( v_s = 3.5 \, \text{m/s} \)? | ||
| 10 | Set up equation with \( v_s = 3.5 \, \text{m/s} \):
\( 0.085 t^2 – 3.5 t + 40.0 = 0 \) |
Concluded that the student cannot catch the bus. |
| (f) Minimum speed to catch the bus and corresponding time and distance. | ||
| 11 | Set discriminant \( D = 0 \) to find minimum speed \( v_{s_{\text{min}}} \):
\( (-v_{s_{\text{min}}})^2 – 4(0.085)(40.0) = 0 \) |
Found the minimum speed required. |
| 12 | Calculate time and distance at \( v_{s_{\text{min}}} \):
\( t = \dfrac{v_{s_{\text{min}}}}{2a_b} = \dfrac{3.692}{2 \times 0.170} \approx 10.86 \, \text{s} \) |
Determined time and distance to catch the bus at minimum speed. |
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A projectile is launched at \( 25 \) \( \text{m/s} \) at an angle of \( 37^{\circ} \). It lands on a platform that is \( 5.0 \) \( \text{m} \) above the launch height.
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(a) The student must run for approximately \( 9.55 \, \text{s} \) and cover \( 47.77 \, \text{m} \).
(b) When she reaches the bus, it is traveling at \( 1.62 \, \text{m/s} \).
(c) **Graph Description**:
– Student’s Path: A straight line with constant slope at \( 5.0 \, \text{m/s} \).
– Bus’s Path: A parabola starting at \( 40.0 \, \text{m} \) with increasing slope.
(d) Second solution \( t \approx 49.27 \, \text{s} \) represents when the bus overtakes the student again. Bus speed at that time: \( 8.38 \, \text{m/s} \).
(e) If the student’s top speed is \( 3.5 \, \text{m/s} \), she will not catch the bus (no real solution, \( D < 0 \)).
(f) Minimum speed to catch the bus: \( 3.69 \, \text{m/s} \), time \( 21.72 \, \text{s} \), covering \( 80.17 \, \text{m} \).
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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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