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Unit 1 - Vectors and Kinematics

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1. Maximum Height Reached by the Coin

Step Formula Derivation Reasoning
1 [katex]v^2 = u^2 + 2as[/katex] Kinematic equation for motion, with [katex]v[/katex] as final velocity, [katex]u[/katex] as initial velocity, [katex]a[/katex] as acceleration, and [katex]s[/katex] as displacement.
2 [katex]0 = (11, \text{m/s})^2 – 2 \times 9.81, \text{m/s}^2 \times s[/katex] At maximum height, final velocity [katex]v = 0[/katex], initial velocity [katex]u = 11, \text{m/s}[/katex] upward, acceleration [katex]a = -9.81, \text{m/s}^2[/katex] (gravity acts downward).
3 Solve for [katex]s[/katex] Calculate the displacement [katex]s[/katex].

2. Position 4.20 Seconds After Being Released

Step Formula Derivation Reasoning
1 [katex]s = ut + \frac{1}{2}at^2[/katex] Kinematic equation for displacement.
2 [katex]s = 11, \text{m/s} \times 4.20, \text{s} – \frac{1}{2} \times 9.81, \text{m/s}^2 \times (4.20, \text{s})^2[/katex] Substitute values for [katex]u[/katex], [katex]a[/katex], and [katex]t[/katex] (time after release).

3. Velocity 4.20 Seconds After Being Released

Step Formula Derivation Reasoning
1 [katex]v = u + at[/katex] Kinematic equation for velocity.
2 [katex]v = 11, \text{m/s} – 9.81, \text{m/s}^2 \times 4.20, \text{s}[/katex] Substitute values for [katex]u[/katex], [katex]a[/katex], and [katex]t[/katex].

4. Time Before Coin Hits the Ground

Step Formula Derivation Reasoning
1 [katex]s = ut + \frac{1}{2}at^2[/katex] Kinematic equation for displacement.
2 [katex]-250, \text{m} = 11, \text{m/s} \times t – \frac{1}{2} \times 9.81, \text{m/s}^2 \times t^2[/katex] Substitute values for [katex]s[/katex] (downward displacement), [katex]u[/katex], and [katex]a[/katex]. Solve for [katex]t[/katex].

Let’s perform the calculations for each part.

The results for each part are as follows:

  1. Maximum height reached by the coin: [katex]\boxed{256.17, \text{meters}}[/katex]
  2. Position of the coin 4.20 seconds after being released: [katex]\boxed{209.68, \text{meters}}[/katex] above the ground
  3. Velocity of the coin 4.20 seconds after being released: [katex]\boxed{-30.20, \text{m/s}}[/katex] (downward, as indicated by the negative sign)
  4. Time before the coin hits the ground: [katex]\boxed{8.35, \text{seconds}}[/katex]

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  1. 256.2 m
  2. 210 m
  3. -30.16 m/s
  4. 8.4 s

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Download Equation Sheet
KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque 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 MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[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
VariableSI 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]
VariableDerived 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. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. 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]

  3. 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]

  4. 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]

  1. 1. Some answers may vary by 1% due to rounding.
  2. Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
  3. Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
  4. Bookmark questions you can’t solve to revisit them later
  5. 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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