| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[v_{\text{rel}} = v_{m} – v_{c} = 30 – 20 = 10\;\text{m/s}\] | The relative speed is the motorcycle’s speed minus the car’s speed. |
| 2 | \[v_{\text{rel}} > 0\] | A positive relative speed means the motorcycle is closing the gap. |
| 3 | \[\boxed{\text{Yes}}\] | Because the motorcycle is faster, it will eventually catch the car. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[t = \frac{\Delta x_{\text{initial}}}{v_{\text{rel}}} = \frac{30}{10} = 3\;\text{s}\] | Time required equals initial separation divided by relative speed. |
| 2 | \[\Delta x = v_{c}\, t = 20 \times 3 = 60\;\text{m}\] | Distance from the car’s start is its speed times the catch-up time. |
| 3 | \[\boxed{t = 3\;\text{s},\; \Delta x = 60\;\text{m}}\] | Final answers for part (b). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[x_{c}(2) = 20 \times 2 = 40\;\text{m}\] | Car’s position after the first \(2\) seconds. |
| 2 | \[x_{m}(2) = 30 \times 2 – 30 = 30\;\text{m}\] | Motorcycle’s position after \(2\) seconds, relative to the car’s starting position. |
| 3 | \[\Delta x_{0} = x_{m} – x_{c} = -10\;\text{m}\] | So the motorcycle is still \(10 ~\text{m}\) behind when acceleration starts (at the \(2\) second mark). So let’s make an equation, for each vehicle, to see if and where they will meet. |
| 4 | \[x_{c} = 40 + 20 t_{1} + \tfrac{1}{2}(1) t_{1}^{2}\] | Car’s position for time \(t_{1}\) seconds after the \(2\) second mark. The equation shows that the car accelerates from the \(40\) meter mark. |
| 5 | \[x_{m} = 30 + 30 t_{1}\] | Motorcycle’s position for time \(t_{1}\) seconds after the \(2\) second mark. The equation shows the motorcycle continues at constant speed from its \(30\) meter mark. |
| 6 | \[x_{m}=x_{c}\;\Rightarrow\;-10 + 10 t_{1} – 0.5 t_{1}^{2}=0\] | We want to find when and if the vehicles meet. So set position equations, from step 4 and 5, equal to each other to find the catch-up (meet-up) time. |
| 7 | \[t_{1}^{2}-20 t_{1}+20 = 0\] | Multiply both sides by \(-2\) to simplify and rearrange to a quadratic equation. |
| 8 | \[t_{1}=\frac{20 \pm \sqrt{400-80}}{2}=1.06\;\text{s}\;\text{(smaller root)}\] | The larger root occurs after the car overtakes; only the smaller root is physical. |
| 9 | \[t = 2 + t_{1} = 2 + 1.06 = 3.06\;\text{s}\] | Total time from the initial start. |
| 10 | \[\Delta x = 40 + 20(1.06) + 0.5(1.06)^{2} \approx 61.7\;\text{m}\] | Distance from the car’s starting point where they meet. |
| 11 | \[\boxed{\text{Yes},\; t = 3.06\;\text{s},\; \Delta x = 61.7\;\text{m}}\] | The motorcycle still catches the car despite the acceleration. |
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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?
A skateboarder, with an initial speed of \( 20.0 \, \text{m/s} \), rolls to the end of friction-free incline of length \( 25 \, \text{m} \). At what angle is the incline oriented above the horizontal?
An object of unknown mass is acted upon by multiple forces:
The coefficients of friction are \(\mu_s = 0.6\) and \(\mu_k = 0.2\). Starting from rest, the object travels \(10 \, \text{m}\) in \(4.5 \, \text{s}\). What is the mass of the unknown object?
Divers in Acapulco jump from a cliff that is \( 36 \, \text{m} \) above the water with an initial vertical velocity of \( 2 \, \text{m/s} \).
The motion of a particle is described in the velocity vs. time graph shown above. Over the nine-second interval shown, we can say that the speed of the particle…
A ball is tossed straight up while the thrower is standing in a moving train car that is moving at a constant velocity. Neglecting air resistance, what is the path of the ball relative to the ground outside the train?
A car moving at 30 m/s makes a head-on collision with a stone wall. From what height would the car have to fall in order to make an equally hard collision with the ground?
Priscilla the Penguin stands at the edge of a rock ledge and tosses a small ice cube directly upward with an initial velocity of \( v_0 \). The ice cube’s initial height above the ground is \( 3.25 \, \text{m} \), and it reaches its maximum height above the ground \( 0.586 \, \text{s} \) after being thrown. The ice cube then plummets to the ground, missing the edge of the rock ledge on its way down.
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 \(10 \, \text{kg}\) box is pushed to the right by an unknown force at an angle of \(25^\circ\) below the horizontal while a friction force of \(50 \, \text{N}\) acts on the box as well. The box accelerates from rest and travels a distance of \(4 \, \text{m}\) where it is moving at \(3 \, \text{m/s}\).
\( \text{Yes} \)
\( t = 3\,\text{s},\; \Delta x = 60\,\text{m} \)
\( \text{Yes},\; t = 3.06\,\text{s},\; \Delta x = 61.7\,\text{m} \text{from the car’s start}\)
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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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