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# Part (a) Determine the sprinter’s constant acceleration during the first \(2 \, \text{seconds}\).
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]d_1 = 100 \, \text{m} \, – \, 90 \, \text{m} = 10 \, \text{m}[/katex] | The first part of the sprint covers 10 meters. |
| 2 | [katex]d_1 = \frac{1}{2} a t_1^2[/katex] | Use the formula for distance under constant acceleration starting from rest: [katex]d = \frac{1}{2} a t^2[/katex]. |
| 3 | [katex]10 \, \text{m} = \frac{1}{2} a (2 \, \text{s})^2 [/katex] | Substitute [katex] d_1 = 10 \, \text{m} [/katex] and [katex] t_1 = 2 \, \text{s} [/katex]. |
| 4 | [katex]10 \, \text{m} = 2 a \, \text{s}^2 [/katex] | Simplify the equation. |
| 5 | [katex]a = 5 \, \text{m/s}^2 [/katex] | Solving for acceleration gives [katex]a[/katex]. |
| 6 | [katex]a = 5 \, \text{m/s}^2[/katex] | Constant acceleration value. |
# Part (b) Determine the sprinter’s velocity after 2 seconds have elapsed.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v = a t_1 [/katex] | Using the formula for velocity under constant acceleration: [katex]v = a t[/katex]. |
| 2 | [katex]v = 5 \, \text{m/s}^2 \times 2 \, \text{s}[/katex] | Substitute [katex] a = 5 \, \text{m/s}^2 [/katex] and [katex] t_1 = 2 \, \text{s} [/katex]. |
| 3 | [katex]v = 10 \, \text{m/s}[/katex] | Solve for [katex]v[/katex]. |
# Part (c) Determine the total time needed to run the full 100 meters.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v = d_2 / t_2 [/katex] | The velocity [katex]v[/katex] is constant for the remaining part of the race. |
| 2 | [katex]10 \, \text{m/s} = 90 \, \text{m} / t_2 [/katex] | Substitute [katex] v = 10 \, \text{m/s} [/katex] and [katex] d_2 = 90 \, \text{m} [/katex]. |
| 3 | [katex]t_2 = 90 \, \text{m} / 10 \, \text{m/s} [/katex] | Rearrange to solve for [katex] t_2 [/katex]. |
| 4 | [katex]t_2 = 9 \, \text{s} [/katex] | Solve for [katex] t_2 [/katex]. |
| 5 | [katex]t_{\text{total}} = t_1 + t_2 = 2 \, \text{s} + 9 \, \text{s} [/katex] | The total time is the sum of the two intervals. |
| 6 | [katex]t_{\text{total}} = 11 \, \text{s} [/katex] | Total time to run 100 meters. |
# Part (d) Draw the displacement vs time curve for the sprinter.
The displacement vs. time graph would show a parabolic curve for the first 2 seconds and a linear relationship thereafter to indicate constant velocity:
1. From [katex] t = 0 [/katex] to [katex] t = 2 [/katex] seconds, the curve will be a parabola opening upwards.
2. From [katex] t = 2[/katex] seconds to [katex] t = 11 [/katex] seconds, the curve will be a straight line with a constant slope of [katex]10 \, \text{m/s}[/katex].
Just ask: "Help me solve this problem."
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A truck is traveling at 35 m/s when the driver realizes the truck as no breaks. He sees a ramp off the road, inclined at 20°, and decides to go up it to help the truck come to a stop. How far does the truck travel before coming to a stop (assume no friction).
A 135.0 N force is applied to a 30.0 kg box at 42 degree angle to the horizontal. If the force of friction is 85.0, what is the net force and acceleration? If the object starts from rest, how far has it traveled in 3.3 sec?
Two identical metal balls are being held side by side at the top of a ramp. Alex lets one ball, \( A \), start rolling down the hill. A few seconds later, Alex’s partner, Bob, starts the second ball, \( B \), down the hill by giving it a push. Ball \( B \) rolls down the hill along a line parallel to the path of the first ball and passes it. At the instant ball \( B \) passes ball \( A \):
Which of the following statements about the acceleration due to gravity is TRUE?
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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
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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