| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (a) Accelerarion of the particle when its displacement is 6 m | ||
| 1 | \[F = ma\] | Newton’s second law relates force \( F \), mass \( m \), and acceleration \( a \). |
| 2 | \[a = \frac{F}{m}\] | Rearrange the formula to solve for acceleration. |
| 3 | \[a = \frac{4\, \text{N}}{0.20\, \text{kg}}\] | Substitute the force from the graph (4 N) and the mass (0.20 kg). |
| 4 | \[a = 20\, \text{m/s}^2\] | Calculate the acceleration. |
| (b) Time taken for the object to be displaced the first 12 m | ||
| 1 | \[\Delta x = v_i t + \frac{1}{2} a t^2\] | Using the kinematic equation with initial velocity \( v_i = 0 \). |
| 2 | \[12 = \frac{1}{2} \cdot 20 \cdot t^2\] | Substitute \( \Delta x = 12 \) m and \( a = 20 \text{ m/s}^2 \). |
| 3 | \[12 = 10 t^2\] | Simplify the equation. |
| 4 | \[t^2 = 1.2\] | Divide both sides by 10. |
| 5 | \[t = \sqrt{1.2}\] | Solve for \( t \). |
| 6 | \[t \approx 1.095\, \text{s}\] | Calculate the time taken. |
| (c) The amount of work done by the net force in displacing the object the first 12 m | ||
| 1 | \[W = F \Delta x\] | Work done \( W \) is the product of force and displacement. |
| 2 | \[W = 4 \times 12\] | Substitute \( F = 4 \text{ N} \) and \( \Delta x = 12 \text{ m} \). |
| 3 | \[W = 48 \text{ J}\] | Calculate the work done. |
| (d) The speed of the object at displacement \( x = 12 \text{ m} \) | ||
| 1 | \[v_x^2 = v_i^2 + 2a \Delta x\] | Use the kinematic equation with initial velocity \( v_i = 0 \). |
| 2 | \[v_x^2 = 0 + 2 \cdot 20 \cdot 12\] | Substitute \( a = 20 \text{ m/s}^2 \) and \( \Delta x = 12 \text{ m} \). |
| 3 | \[v_x^2 = 480\] | Calculate \( v_x^2 \). |
| 4 | \[v_x = \sqrt{480}\] | Solve for \( v_x \). |
| 5 | \[v_x \approx 21.9 \, \text{m/s}\] | Calculate the velocity. |
| (e) The final speed of the object at displacement \( x = 20 \text{ m} \) | ||
| 1 | \[W_{total} = W_{1} + W_{2}\] | Calculate total work done by summing areas under the \( F \) vs. \( x \) graph. |
| 2 | \[W_{1} = F_{1} \times \Delta x_{1} = 4 \times 12 = 48 \, \text{J}\] | The work done on the first section (rectangle 0 to 12 m). |
| 3 | \[W_{2} = \frac{1}{2} \cdot 4 \cdot 8 = 16 \, \text{J}\] | The work done on the second section (triangular area from 12 m to 20 m). |
| 4 | \[W_{total} = 48 + 16 = 64 \, \text{J}\] | Total work done. |
| 5 | \[\text{K.E.} = \frac{1}{2}m v_x^2\] | Relate total work done to kinetic energy gain. |
| 6 | \[64 = \frac{1}{2} \cdot 0.20 \cdot v_x^2\] | Substitute \( m = 0.20 \, \text{kg} \). |
| 7 | \[v_x^2 = 640\] | Solve for \( v_x^2 \). |
| 8 | \[v_x = \sqrt{640}\] | Solve for \( v_x \). |
| 9 | \[\boxed{v_x \approx 25.3 \, \text{m/s}}\] | Calculate the final speed at \( x = 20 \text{ m} \). |
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A \(2 \, \text{kg}\) model rocket is launched with a thrust force of \(275 \, \text{N}\) and reaches a height of \(90 \, \text{m}\), at which point the thrust cuts out, but the rocket continues moving at \(150 \, \text{m/s}\). What is the average air resistance force acting on the rocket during its ascent?
Two objects attract each other gravitationally with a force of \( 2.5 \times 10^{-10} \) \( \text{N} \) when they are \( 0.25 \) \( \text{m} \) apart. Their total mass is \( 4.00 \) \( \text{kg} \). Find their individual masses.
A car accelerates uniformly from rest to \( 29.4 \) m/s in \( 6.93 \) s along a level stretch of road. Ignoring friction, determine the average power in both watts and horsepower (\( 1 \text{ horsepower} = 745.7 \text{ Watts} \)) required to accelerate the car if:
The graph above represents the motion of an object traveling in a straight line as a function of time. What is the average speed of the object during the first four seconds?
A sled glides across ice and eventually stops. This stopping is best explained by ____.
When a golf ball is dropped to the pavement, it bounces back up.
A space probe far from the Earth is traveling at 14.8 km/s. It has mass 1312 kg. The probe fires its rockets to give a constant thrust of 156 kN for 220 seconds. It accelerates in the same direction as its initial velocity. In this time it burns 150 kg of fuel. Calculate final speed of the space probe in km/s.
Note: This is a bonus question. Skip if you haven’t yet taken calculus.
A uniform solid cylinder of mass \( M \) and radius \( R \) is initially at rest on a frictionless horizontal surface. A massless string is attached to the cylinder and is wrapped around it. The string is then pulled with a constant force \( F \) , causing the cylinder to rotate about its center of mass. After the cylinder has rotated through an angle \( \theta \), what is the kinetic energy of the cylinder in terms of \( F \) and \( \theta \)?
A boulder is raised above the ground so that its potential energy is \(550 \, \text{J}\). Then it is dropped. Assuming \(92 \, \text{J}\) of energy was lost to air resistance, what is the kinetic energy of the boulder just before it hits the ground?
A 0.035 kg bullet moving horizontally at 350 m/s embeds itself into an initially stationary 0.55 kg block. Air resistance is negligible.
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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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