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Step | Derivation/Formula | Reasoning |
---|---|---|

(a) Time and distance for the student to overtake the bus. |
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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. |
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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. |
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8 | Description of the graph:
– |
Visual representation of positions over time. |

(d) Significance of the second time solution and bus’s speed at that point. |
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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 \( v_s = 3.5 \, \text{m/s} \)? |
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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. |
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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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- Statistics

Intermediate

Mathematical

GQ

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?

- 1D Kinematics

Intermediate

Mathematical

MCQ

- Motion Graphs

Beginner

Conceptual

MCQ

- 1D Kinematics

Intermediate

Mathematical

MCQ

A cart starts from rest and accelerates uniformly at 4.0 m/s^{2} for 5.0 s. It next maintains the velocity it has reached for 10 s. Then it slows down at a steady rate of 2.0 m/s^{2} for 4.0 s. What is the final speed of the car?

- 1D Kinematics

Intermediate

Mathematical

GQ

Traveling at a speed of 15.9 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.659. What is the speed of the automobile after 1.59 s have elapsed? Ignore the effects of air resistance.

- 1D Kinematics, Linear Forces

(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] |

General Metric Conversion Chart

Conversion Example

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] | 0.000000000001 |

Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |

Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |

Milli- | m | [katex]10^{-3}[/katex] | 0.001 |

Centi- | c | [katex]10^{-2}[/katex] | 0.01 |

Deci- | d | [katex]10^{-1}[/katex] | 0.1 |

(Base unit) | – | [katex]10^{0}[/katex] | 1 |

Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |

Hecto- | h | [katex]10^{2}[/katex] | 100 |

Kilo- | k | [katex]10^{3}[/katex] | 1,000 |

Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |

Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |

Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |

- 1. Some answers may vary by 1% due to rounding.
- Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
- Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
- Bookmark questions you can’t solve to revisit them later
- 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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