| 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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The figure shows the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the displacement of the player…
Two objects are dropped from rest from the same height. Object \( A \) falls through a distance \( d_A \) during a time \( t \), and object \( B \) falls through a distance \( d_B \) during a time \( 2t \). If air resistance is negligible, what is the relationship between \( d_A \) and \( d_B \)?
A cart is initially moving at 0.5 m/s along a track. The cart comes to rest after traveling 1 m. The experiment is repeated on the same track, but now the cart is initially moving at 1 m/s. How far does the cart travel before coming to rest?
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?
A horizontal spring with spring constant 162 N/m is compressed 50 cm and used to launch a 3 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the rough surface is 0.2. Find the total distance the box travels before stopping.
A baseball is thrown vertically into the air with a velocity \( v \), and reaches a maximum height \( h \). At what height was the baseball moving with one-half its original velocity? Assume air resistance is negligible.
A rocket, initially at rest, is fired vertically upward with an acceleration of \( 12.0 \, \text{m/s}^2 \). At an altitude of \( 1.00 \, \text{km} \), the rocket engine cuts off. Drag is negligible.
A cart with an initial velocity of \(5.0 ~ \text{m/s}\)to the right experiences a constant acceleration of \(2.0 ~ \text{m/s}^2\) to the right. What is the cart’s displacement during the first \(6.0 ~ \text{s}\) of this motion?
A gun can fire a bullet to height \( h \) when fired straight up. If the same gun is pointed at an angle of \( 45^\circ \) from the vertical, what is the new maximum height of the projectile?
The International Space Station travels at \( 7660 \, \text{m/s} \). Find the average velocity of the space station if it takes \( 90 \, \text{minutes} \) to make one full orbit around Earth.
(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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