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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( d_1 = v_i t + \frac{1}{2} a t^2 \) | Using the kinematic equation for distance travelled \( d \) with initial velocity \( v_i = 0 \), time \( t \), and constant acceleration \( a \). |
| 2 | \( 5 \, \text{m} = \frac{1}{2} a (1 \, \text{s})^2 \) | The distance travelled in the first second is given as \( 5 \, \text{m} \). Substitute \( t = 1 \, \text{s} \) to find the acceleration. |
| 3 | \( 5 = \frac{1}{2} a \) | Simplifying the equation from Step 2. \( 1^2 \) is still \( 1 \), so it simplifies to \( 5 = \frac{1}{2} a \). |
| 4 | \( a = 10 \, \text{m/s}^2 \) | Solving for \( a \) by multiplying both sides of the equation by 2. |
| 5 | \( d_2 = v_i t + \frac{1}{2} a (t + 1)^2 \) | Using the distance formula to calculate the total distance travelled in the first two seconds. Initial velocity \( v_i = 0 \), distance for second time interval \( t_2 = 1 \,\text{s} \). |
| 6 | \( d_2 – d_1 = (v_i + a \cdot 1) \cdot 1 + \frac{1}{2} a \cdot 1^2 \) | Subtract the distance travelled in the first second from the distance travelled in the first two seconds to isolate the distance covered in the second time interval. |
| 7 | \( 15 \, \text{m} – 5 \, \text{m} = 5 + \frac{1}{2} a \cdot 1^2 \) | Given that the distance travelled in the next second is \( 15 \, \text{m} \), the distance covered in the second interval would be \( 15 \, \text{m} – 5 \, \text{m} = 10 \, \text{m} \). |
| 8 | \( 5 + \frac{1}{2} a = 10 \, \text{m} \) | Simplify the above equation \( \frac{1}{2} a \cdot 1 = 10 – 5 = 5 \, \text{m} \). |
| 9 | \( \frac{1}{2} a = 5 \, \text{m} \) | By isolating \( a \) and solving helps in verifying the correctness of previous values calculated. |
| 10 | \( d_3 = \frac{1}{2} a (3)^2 \) | Finally, calculate the total distance travelled in the first three seconds. Using \( t = 3 \, \text{s}\) while keeping other variables the same. |
| 11 | \( d_3 = \frac{1}{2} \cdot 10 \, \text{m/s}^2 \cdot 9 \, \text{s}^2 \) | Substitute the known values for \( a \) and \( t \) into the equation for \( d_3 \). |
| 12 | \( d_3 = 5 \cdot 9 \, \text{m} \) | Simplify by multiplying \( \frac{1}{2} \cdot 10 \) to get \( 5 \) and \( 3^2 = 9 \). |
| 13 | \( d_3 = 45 \, \text{m} \) | The total distance travelled after the 3 seconds is \( 45 \, \text{m} \). |
So, the correct answer is \(\boxed{45 \, \text{m}}\), which corresponds to option (e).
Just ask: "Help me solve this problem."
A ranger in a national park is driving at \( 56 \, \text{km/h} \) when a deer jumps onto the road \( 65 \, \text{m} \) ahead of the vehicle. After a reaction time of \( t \, \text{s} \), the ranger applies the brakes to produce an acceleration of \( -3 \, \text{m/s}^2 \). What is the maximum reaction time allowed if the ranger is to avoid hitting the deer?
Which of the following statements about the acceleration due to gravity is TRUE?
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 projectile of mass 0.750 kg is shot straight up with an initial speed of 18.0 m/s.
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.
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) | Â |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| Â | \(W = Fd \cos\theta\) |
| 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
Example of using unit analysis: Convert 5 kilometers to millimeters.Â
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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