# Part (a) Determine the sprinter’s constant acceleration during the first \(2 \, \text{seconds}\).
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(d_1 = 100 \, \text{m} \, – \, 90 \, \text{m} = 10 \, \text{m}\) | The first part of the sprint covers 10 meters. |
| 2 | \(d_1 = \frac{1}{2} a t_1^2\) | Use the formula for distance under constant acceleration starting from rest: \(d = \frac{1}{2} a t^2\). |
| 3 | \(10 \, \text{m} = \frac{1}{2} a (2 \, \text{s})^2 \) | Substitute \( d_1 = 10 \, \text{m} \) and \( t_1 = 2 \, \text{s} \). |
| 4 | \(10 \, \text{m} = 2 a \, \text{s}^2 \) | Simplify the equation. |
| 5 | \(a = 5 \, \text{m/s}^2 \) | Solving for acceleration gives \(a\). |
| 6 | \(a = 5 \, \text{m/s}^2\) | Constant acceleration value. |
# Part (b) Determine the sprinter’s velocity after 2 seconds have elapsed.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v = a t_1 \) | Using the formula for velocity under constant acceleration: \(v = a t\). |
| 2 | \(v = 5 \, \text{m/s}^2 \times 2 \, \text{s}\) | Substitute \( a = 5 \, \text{m/s}^2 \) and \( t_1 = 2 \, \text{s} \). |
| 3 | \(v = 10 \, \text{m/s}\) | Solve for \(v\). |
# Part (c) Determine the total time needed to run the full 100 meters.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v = d_2 / t_2 \) | The velocity \(v\) is constant for the remaining part of the race. |
| 2 | \(10 \, \text{m/s} = 90 \, \text{m} / t_2 \) | Substitute \( v = 10 \, \text{m/s} \) and \( d_2 = 90 \, \text{m} \). |
| 3 | \(t_2 = 90 \, \text{m} / 10 \, \text{m/s} \) | Rearrange to solve for \( t_2 \). |
| 4 | \(t_2 = 9 \, \text{s} \) | Solve for \( t_2 \). |
| 5 | \(t_{\text{total}} = t_1 + t_2 = 2 \, \text{s} + 9 \, \text{s} \) | The total time is the sum of the two intervals. |
| 6 | \(t_{\text{total}} = 11 \, \text{s} \) | 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 \( t = 0 \) to \( t = 2 \) seconds, the curve will be a parabola opening upwards.
2. From \( t = 2\) seconds to \( t = 11 \) seconds, the curve will be a straight line with a constant slope of \(10 \, \text{m/s}\).
A Major Upgrade To Phy Is Coming Soon — Stay Tuned
We'll help clarify entire units in one hour or less — guaranteed.
A self paced course with videos, problems sets, and everything you need to get a 5. Trusted by over 15k students and over 200 schools.
A whiffle ball is tossed straight up, reaches a highest point, and falls back down. Air resistance is not negligible. Which of the following statements are true?
A ball is thrown straight up. What are the velocity and acceleration of the ball at the highest point in its path?
Toy car W travels across a horizontal surface with an acceleration of \( a_w \) after starting from rest. Toy car Z travels across the same surface toward car W with an acceleration of \( a_z \), after starting from rest. Car W is separated from car Z by a distance \( d \). Which of the following pairs of equations could be used to determine the location on the horizontal surface where the two cars will meet, and why?
The alarm at a fire station rings and a 79.34-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 4.20 m). Just before landing, his speed is 1.36 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?
Can an object’s average velocity equal zero when object’s speed is greater than zero? Explain using the formula for average velocity vs speed.
A ball is thrown straight up with a speed of \( 30 \) \( \text{m/s} \), and air resistance is negligible.
Given the graph of velocity versus time for a duck flying due south for the winter, at what labeled point did the duck stop its forward motion?
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 stone is thrown vertically upwards with a speed of \( 20.0 \) \( \text{m/s} \). How fast is it moving when it reaches a height of \( 12.0 \) \( \text{m} \)?
A block starts from rest at the top of a \(50^\circ\) incline. The coefficient of kinetic friction between the block and the incline is \(0.4\). If the block reaches a velocity of \(7 \, \text{m/s}\) at the bottom of the incline, what is the length of the incline?
By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
One price to unlock most advanced version of Phy across all our tools.
per month
Billed Monthly. Cancel Anytime.
Try our free calculator to see what you need to get a 5 on the 2026 AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed
