Kinematics Study Notes: Distance, Displacement, Speed, Velocity, Reference Frames, and Motion Graphs

DISTANCE VS DISPLACEMENT

• Distance: total length of the path traveled by an object; scalar quantity with magnitude only.
• Displacement: straight-line distance from the initial position to the final position, with a direction toward the final position; a vector quantity.
• Example excerpt (Page 3): Distance Traveled: 695 m; Displacement: 0 m (Saitama Park). Interpretation: the object ends up at its starting point, so net displacement is zero though distance traveled is 695 m.

SPEED VS VELOCITY

• Speed: scalar quantity; magnitude only; rate of motion along the path.
• Velocity: vector quantity; magnitude and direction; rate of change of position.
• Units: both have SI unit of $1\frac{m}{s}$; typically expressed in m/s with a direction for velocity.

INSTANTANEOUS VS AVERAGE SPEED

• Instantaneous speed: speed at a particular moment in time.
• Average speed: total distance traveled divided by total time elapsed.

VELOCITY AND ACCELERATION AS FUNCTIONS OF TIME (DERIVATIVES)

• Position is denoted by $s(t)$.
• Velocity: $v(t) = s'(t) = \dfrac{ds}{dt}$; rate at which position changes.
• Acceleration: $a(t) = s''(t) = \dfrac{d^2s}{dt^2}$; derivative of velocity.
• Example:
  • Given displacement $s(t) = 5t^3 + 3t + 8$ where t is in seconds.
  • Velocity: $v(t) = \dfrac{ds}{dt} = 15t^2 + 3$
  • Acceleration: $a(t) = \dfrac{d^2s}{dt^2} = 30t$
• Evaluations:
  • A) Velocity at $t = 3\ \text{s}$: $v(3) = 15(3)^2 + 3 = 15\cdot 9 + 3 = 135 + 3 = 138\ \text{m/s}$
  • B) Acceleration at $t = 3\ \text{s}$: $a(3) = 30\cdot 3 = 90\ \text{m/s}^2$
  • C) When will velocity exceed $200\ \text{m/s}$?
  • Solve: 15t^2 + 3 > 200
  • t^2 > \dfrac{200-3}{15} = \dfrac{197}{15} \approx 13.1333
  • t > \sqrt{\dfrac{197}{15}} \approx 3.63\ \text{s}

MOTION GRAPHS (GENERAL IDEAS)

• Is Mumen Rider accelerating? (conceptual question about interpreting acceleration from graphs.)
• Types:
  • Straight line vs curved line on a position-time graph.
  • Constant slope vs changing slope (constant velocity vs changing velocity).
  • Constant motion vs changing motion; Not accelerating vs accelerating.

SLOPE AND AREAS ON GRAPHS

• Position-time (P-t) graph:
  • Slope of the graph (tangent) at time t gives the instantaneous velocity: $v(t) = \left.\dfrac{ds}{dt}\right|_{t}\,.$
  • Slope of the secant line between two times gives the average velocity: $\bar{v} = \dfrac{s(t<em>2) - s(t</em>1)}{t<em>2 - t</em>1}.$
  • The slope of a tangent line is difficult to obtain exactly; approximate by using two nearby points (e.g., t = 2.9 s and t = 3.1 s).
  • Note: The area under a Position-time graph is not a standard physical quantity like displacement or distance and is generally not meaningful for kinematics purposes.
• Velocity-time (V-t) graph:
  • Slope of the graph represents acceleration: $a(t) = \dfrac{dv}{dt}.$
  • Area under the V-t graph represents displacement: $d = \int v\,dt.$
• Acceleration-time (A-t) graph:
  • Slope represents jerk (the rate of change of acceleration).
  • Area under the A-t graph represents the change in velocity: $\Delta v = \int a\,dt.$

POSITION-TIME GRAPH (KEY TAKEAWAYS)

• The slope of the tangent line at time t gives the instantaneous velocity: $v<em>{inst}(t) = \left.\dfrac{ds}{dt}\right|</em>{t}.$
• The slope of the secant line between two points gives the average velocity: $v_{avg} = \dfrac{\Delta s}{\Delta t}.$
• To approximate the tangent slope, you can use two nearby time points (e.g., t = 2.9 s and t = 3.1 s).

PRACTICAL NOTES ON GRAPHS

• If a velocity-time graph is flat (horizontal), acceleration is zero (not accelerating).
• If a velocity-time graph is a line with positive slope, acceleration is constant and positive.
• If the velocity-time graph is a curved line, acceleration is changing (not constant).

NEXT TOPIC

• Free fall (gravity-driven motion) will be studied next.

REFERENCE FRAMES (CONCEPTS AND EXAMPLES)

• A reference frame is an abstract coordinate system plus a set of fixed points used to fix the coordinate system and standardize measurements within that frame.
• Perception exercise (Page 9): Looking out of a bus window, how do you perceive moving objects (trees, cars) relative to your frame of reference? Different observers (on bus vs. standing still) observe different velocities for the same object.
• Example reference speeds (Page 11):
  • Saitama: 0 m/s (stationary frame reference)
  • Running man: 30 m/s (direction specified, e.g., E)
  • Bus passing by: 45 m/s (direction specified, e.g., W)

ADDITIONAL PRACTICAL POINTS ABOUT DERIVATIVES AND TIME-DEPENDENT MOTION

• If position is given as a function of time, you can obtain velocity by differentiation and acceleration by differentiation again:
  • $v(t) = \dfrac{ds}{dt},\quad a(t) = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}.$
• Time-dependent problems often use derivatives to find velocity/acceleration, and integrals to relate velocity and displacement (e.g., displacement as the integral of velocity over time).

EQUATIONS TO REMEMBER (LAtex)

• Position, velocity, acceleration:
  • $v(t) = \dfrac{ds}{dt}, \quad a(t) = \dfrac{d^2s}{dt^2}.$
• Example: if $s(t) = 5t^3 + 3t + 8$, then
  • $v(t) = \dfrac{ds}{dt} = 15t^2 + 3,$
  • $a(t) = \dfrac{d^2s}{dt^2} = 30t.$
• Evaluations:
  • $v(3) = 15(3)^2 + 3 = 138\ \text{m/s},$
  • $a(3) = 30(3) = 90\ \text{m/s}^2.$
• When will velocity exceed 200 m/s?
  • Solve: 15t^2 + 3 > 200\quad\Rightarrow\quad t > \sqrt{\dfrac{197}{15}} \approx 3.63\ \text{s}.