0 attempts
0% avg
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[v_{\text{rel}} = v_{m} – v_{c} = 30 – 20 = 10\;\text{m/s}\] | The relative speed is the motorcycle’s speed minus the car’s speed. |
| 2 | \[v_{\text{rel}} > 0\] | A positive relative speed means the motorcycle is closing the gap. |
| 3 | \[\boxed{\text{Yes}}\] | Because the motorcycle is faster, it will eventually catch the car. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[t = \frac{\Delta x_{\text{initial}}}{v_{\text{rel}}} = \frac{30}{10} = 3\;\text{s}\] | Time required equals initial separation divided by relative speed. |
| 2 | \[\Delta x = v_{c}\, t = 20 \times 3 = 60\;\text{m}\] | Distance from the car’s start is its speed times the catch-up time. |
| 3 | \[\boxed{t = 3\;\text{s},\; \Delta x = 60\;\text{m}}\] | Final answers for part (b). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[x_{c}(2) = 20 \times 2 = 40\;\text{m}\] | Car’s position after the first \(2\) seconds. |
| 2 | \[x_{m}(2) = 30 \times 2 – 30 = 30\;\text{m}\] | Motorcycle’s position after \(2\) seconds, relative to the car’s starting position. |
| 3 | \[\Delta x_{0} = x_{m} – x_{c} = -10\;\text{m}\] | So the motorcycle is still \(10 ~\text{m}\) behind when acceleration starts (at the \(2\) second mark). So let’s make an equation, for each vehicle, to see if and where they will meet. |
| 4 | \[x_{c} = 40 + 20 t_{1} + \tfrac{1}{2}(1) t_{1}^{2}\] | Car’s position for time \(t_{1}\) seconds after the \(2\) second mark. The equation shows that the car accelerates from the \(40\) meter mark. |
| 5 | \[x_{m} = 30 + 30 t_{1}\] | Motorcycle’s position for time \(t_{1}\) seconds after the \(2\) second mark. The equation shows the motorcycle continues at constant speed from its \(30\) meter mark. |
| 6 | \[x_{m}=x_{c}\;\Rightarrow\;-10 + 10 t_{1} – 0.5 t_{1}^{2}=0\] | We want to find when and if the vehicles meet. So set position equations, from step 4 and 5, equal to each other to find the catch-up (meet-up) time. |
| 7 | \[t_{1}^{2}-20 t_{1}+20 = 0\] | Multiply both sides by \(-2\) to simplify and rearrange to a quadratic equation. |
| 8 | \[t_{1}=\frac{20 \pm \sqrt{400-80}}{2}=1.06\;\text{s}\;\text{(smaller root)}\] | The larger root occurs after the car overtakes; only the smaller root is physical. |
| 9 | \[t = 2 + t_{1} = 2 + 1.06 = 3.06\;\text{s}\] | Total time from the initial start. |
| 10 | \[\Delta x = 40 + 20(1.06) + 0.5(1.06)^{2} \approx 61.7\;\text{m}\] | Distance from the car’s starting point where they meet. |
| 11 | \[\boxed{\text{Yes},\; t = 3.06\;\text{s},\; \Delta x = 61.7\;\text{m}}\] | The motorcycle still catches the car despite the acceleration. |
Just ask: "Help me solve this problem."
We'll help clarify entire units in one hour or less — guaranteed.
A rock is thrown vertically upward with a velocity of \( 20 \, \text{m/s} \) from the edge of a bridge \( 42 \, \text{m} \) above a river.
A \(25 \, \text{g}\) steel ball is attached to the top of a \(24 \, \text{cm}\)-diameter vertical wheel of negligible mass. Starting from rest, the wheel accelerates at \(470 \, \text{rad/s}^2\). The ball is released after \(\frac{3}{4}\) of a revolution. How high does it go above the center of the wheel?
A girl, standing still, tosses a ball vertically upwards. One second later, she tosses up another ball at the same velocity. The balls collide \( 0.5 \) \( \text{s} \) after the second ball is tossed. With what velocity were they tossed? The acceleration due to gravity is \( 9.8 \) \( \text{m/s}^2 \).
Which of the following statements about the acceleration due to gravity is TRUE?
Person A throws a ball horizontally from a cliff \( 20 \) \( \text{m} \) tall at \( 12 \) \( \text{m/s} \). Person B is running to the right on the ground and catches the ball at the same height it would’ve landed after running \( 15 \) \( \text{m} \). How fast was Person B running?
A car moves forward at a steady \( 10 \) \( \text{m/s} \) for \( 5 \) \( \text{s} \). The driver slams the brakes and brings it to rest in \( 2 \) \( \text{s} \). Without waiting, the driver immediately accelerates backward (negative velocity) for \( 3 \) \( \text{s} \) until reaching \( 8 \) \( \text{m/s} \) in reverse. Draw the velocity vs. time graph.
A driver is driving at \( 40 \, \text{m/s} \) when the light turns red in front of her. It takes the driver \( 0.9 \, \text{s} \) to react and hit the brakes. After this, the car slows with an acceleration of \( 3.5 \, \text{m/s}^2 \). What is the total distance traveled by the car?
A projectile has the least speed at what point in its path?
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?
\( \text{Yes} \)
\( t = 3\,\text{s},\; \Delta x = 60\,\text{m} \)
\( \text{Yes},\; t = 3.06\,\text{s},\; \Delta x = 61.7\,\text{m} \text{from the car’s start}\)
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
