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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (a) How much time does it take for a diver to reach the water? | ||
| 1 | Use the vertical motion equation:
\( y = y_0 + v_{0y} t + \dfrac{1}{2} a t^2 \) |
Relates vertical position, initial velocity, time, and acceleration. |
| 2 | Set final position \( y = 0 \, \text{m} \), initial position \( y_0 = 36 \, \text{m} \), initial velocity \( v_{0y} = +2 \, \text{m/s} \), and acceleration \( a = -9.8 \, \text{m/s}^2 \):
\( 0 = 36 + 2 t – 4.9 t^2 \) |
Plugged in known values; acceleration is negative due to gravity. |
| 3 | Rearrange the equation to standard quadratic form:
\( -4.9 t^2 + 2 t + 36 = 0 \) |
Prepared to solve the quadratic equation for \( t \). |
| 4 | Use the quadratic formula \( t = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a} \), where \( a = 4.9 \), \( b = -2 \), \( c = -36 \):
Discriminant \( D = b^2 – 4ac \): |
Calculated the discriminant for the quadratic equation. |
| 5 | Solve for \( t \):
\( t = \dfrac{-b \pm \sqrt{D}}{2a} = \dfrac{-(-2) \pm \sqrt{709.6}}{2 \times 4.9} \) |
Applied quadratic formula to find possible values of \( t \). |
| 6 | Select the positive root (physical solution):
\( t = \dfrac{2 + 26.65}{9.8} \approx \dfrac{28.65}{9.8} \approx 2.92 \, \text{s} \) |
Negative time is not physically meaningful here; chose positive time. |
| (b) What is the diver’s velocity right before they hit the water? | ||
| 7 | Use the velocity equation:
\( v = v_0 + a t \) |
Relates final velocity, initial velocity, acceleration, and time. |
| 8 | Substitute known values \( v_0 = +2 \, \text{m/s} \), \( a = -9.8 \, \text{m/s}^2 \), \( t = 2.92 \, \text{s} \):
\( v = 2 + (-9.8)(2.92) \) |
Calculated final velocity; negative sign indicates downward direction. |
| 9 | Express the speed and direction:
The diver’s speed is \( 26.62 \, \text{m/s} \) downward. |
Provided the magnitude and clarified the direction. |
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A \( 1000 \) \( \text{kg} \) car is traveling east at \( 20 \) \( \text{m/s} \) when it collides perfectly inelastically with a northbound \( 2000 \) \( \text{kg} \) car traveling at \( 15 \) \( \text{m/s} \). If the coefficient of kinetic friction is \( 0.9 \), how far, and at what angle do the two cars skid before coming to a stop?
The graph shows the acceleration as a function of time for an object that is at rest at time \( t = 0 \) \( \text{s} \). The distance traveled by the object between \( 0 \) and \( 2 \) \( \text{s} \) is most nearly
A cart begins to move from rest on a horizontal track. Which of the following correctly indicates the magnitude of the average velocity of the cart during the interval shown and provides a valid explanation?
Hint: when solving this, its consider that the area of the acceleration vs time graph tells you the change in velocity.
A ball rolls down a ramp and gains speed. Its velocity is increasing in the negative direction. What can be said about its acceleration?
A red car, initially at rest, travels east with an acceleration of \( 3.5 \, \text{m/s}^2 \). At the same time as the red car starts to move, a blue car is traveling west at \( 15 \, \text{m/s} \) and accelerating at \( 1.2 \, \text{m/s}^2 \). If they are \( 600 \, \text{m} \) apart the moment the red car starts to move and they are traveling towards each other, where and when will they meet?
Above is the graph of the velocity vs. time of a duck flying due south for the winter. At what point might the duck begin reversing directions?
A stone is thrown vertically upward with a speed of \( 24.0 \) \( \text{m/s} \).
A projectile of mass 0.750 kg is shot straight up with an initial speed of 18.0 m/s.
Person A throws a ball horizontally from a cliff \( 20 \) \( \text{m} \) tall at \( 12 \) \( \text{m/s} \). Person B is running to the right on the ground and catches the ball at the same height it would’ve landed after running \( 15 \) \( \text{m} \). How fast was Person B running?
A ball is dropped from the top of a tall building. At the same instant, a second ball is thrown upward from the ground level. When the two balls pass one another, one on the way up, the other on the way down, compare the magnitudes of their acceleration:
A) \( 2.92 \, \text{s} \)
B) \( -26.62 \, \text{m/s} \)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| 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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