0 attempts
0% avg
We examine each statement using the velocity–time graph and fundamental kinematics.
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| (a) Motion from \(t = 4\,\text{s}\) to \(t = 10\,\text{s}\) | Velocity in this interval is the horizontal segment at \[v = +10\;\text{m\,s}^{-1}\] |
Because the velocity is non–zero and constant, the object is moving at a steady speed. “Not moving” would require \(v = 0\). Statement (a) is false. |
| (b) Initial vs. final position |
Displacement is the signed area under the \(v(t)\) curve.
|
The signed areas do not cancel; a net \(+10\,\text{m}\) remains. Hence the starting and ending positions differ. Statement (b) is false. |
| (c) Speed behaviour from \(t = 14\,\text{s}\) to \(t = 16\,\text{s}\) | Velocity changes from \(v(14)=0\) to \(v(16)=-5\,\text{m\,s}^{-1}\). | The magnitude of velocity rises from \(0\) to \(5\,\text{m\,s}^{-1}\); the object accelerates in the negative direction and its speed increases. It is therefore speeding up, not slowing down. Statement (c) is false. |
| (d) Average acceleration: \(0\!\to\!4\,\text{s}\) vs. \(34\!\to\!36\,\text{s}\) |
\[\begin{aligned} a_{0\!\to\!4}&=\frac{v(4)-v(0)}{4-0}=\frac{10-0}{4}=2.5\,\text{m\,s}^{-2}\\[6pt] a_{34\!\to\!36}&=\frac{v(36)-v(34)}{36-34}=\frac{0-(-5)}{2}=\frac{5}{2}=2.5\,\text{m\,s}^{-2} \end{aligned}\] However, the velocity actually began rising from \(-5\,\text{m\,s}^{-1}\) at \(t=30\,\text{s}\); the slope of that entire straight segment gives the true (constant) acceleration: \[a = \frac{0-(-5)}{36-30}=\frac{5}{6}\,\text{m\,s}^{-2}\approx0.83\,\text{m\,s}^{-2}\] |
The acceleration during \(34\!\to\!36\,\text{s}\) equals the constant slope of the whole 30–36 s segment, \(0.83\,\text{m\,s}^{-2}\), not \(2.5\,\text{m\,s}^{-2}\). Thus \[a_{0\!\to\!4} \neq a_{34\!\to\!36}\] Statement (d) is true. |
Only option D is correct.
Just ask: "Help me solve this problem."
We'll help clarify entire units in one hour or less — guaranteed.
An object of unknown mass is acted upon by multiple forces:
The coefficients of friction are \(\mu_s = 0.6\) and \(\mu_k = 0.2\). Starting from rest, the object travels \(10 \, \text{m}\) in \(4.5 \, \text{s}\). What is the mass of the unknown object?
The International Space Station travels at \( 7660 \, \text{m/s} \). Find the average velocity of the space station if it takes \( 90 \, \text{minutes} \) to make one full orbit around Earth.
Which car controls directly allow the driver to cause acceleration?
Above is a graph of the \(distance\) vs. time for car moving along a road. According the graph, at which of the following times would the automobile have been accelerating positively?
An elevator of height \(h\) ascends with constant acceleration \(a\). When it crosses a platform, it has acquired a velocity \(u\). At this instant a bolt drops from the top of the elevator. Find the time for the bolt to hit the floor of the elevator. Give your answer in terms of \(h\), \(a\), and any constant.
In which of these cases is the rate of change of the particle’s displacement constant?
A cart with an initial velocity of \(5.0 ~ \text{m/s}\)to the right experiences a constant acceleration of \(2.0 ~ \text{m/s}^2\) to the right. What is the cart’s displacement during the first \(6.0 ~ \text{s}\) of this motion?
A block is projected up a ramp with an initial speed \( v_0 \). It travels along the surface of the ramp with constant acceleration \( a \). Take the positive direction of motion to be up the ramp. If the acceleration vector points opposite the initial velocity vector, which of the following MUST be true?
A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s. Then it slows down at a steady rate of 2.0 m/s2 for 4.0 s. What is the final speed of the car?
A spacecraft accelerates at a rate of \(20.0 \, \text{m/s}^2\).
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.
We crafted THE Ultimate A.P Physics 1 Program so you can learn faster and score higher.
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
