A1-2. Motion Graph Gradients and Areas

Core Graphical Principles

In motion graphs, we frequently use the gradient of a line and the area under a line to find the values of physical quantities.

The Gradient: Calculated by choosing two points on the line and dividing the change in the y-axis by the change in the x-axis.

$$\text{Gradient} = \dfrac{\Delta y}{\Delta x}$$

The Area Under the Curve: In IB Physics, you will mostly calculate the area under straight lines by breaking the space into geometric shapes (rectangles and triangles).

Position-Time Graphs

A position-time graph plots how far an object is from a reference point over time.

Gradient = Velocity

Since the y-axis is position ($x$) and the x-axis is time ($t$):

  • Average Velocity: The gradient of a straight line (secant line) connecting two distinct points on the graph represents the average velocity over that time interval.
  • Instantaneous Velocity: The gradient of the tangent line touching the curve at a specific point in time represents the instantaneous velocity.
$$\text{Gradient} = \dfrac{\Delta x}{\Delta t} = v$$

Interpreting Average Velocity on a Curve:

When an object accelerates or decelerates, its position-time graph is curved. We can find the average velocity between any two times by drawing a straight line between those points.

(A) Line AC is steeper than Line AB t (s) x (m) 1 2 3 4 -1 0 1 2 3 4 5 A C B Δt Δx > 0 (B) t (s) x (m) 1 2 3 4 -1 0 1 2 3 4 5 A C B Δt Δx < 0
  • Graph (A): The secant line AC is steeper than AB, showing a higher magnitude of average velocity over that interval. Both segments result in a positive displacement ($\Delta x > 0$).
  • Graph (B): Between points C and B, the displacement is negative ($\Delta x < 0$), yielding a negative average velocity.
  • Horizontal Line: The displacement stays the same. The object is stationary ($v = 0$).
  • Straight Sloped Line: The object is moving with constant velocity.

Instantaneous Velocity & Speed

While average velocity gives the overall rate of displacement over a time interval, instantaneous velocity describes how fast an object is moving, and in what direction, at a precise moment in time.

  • Instantaneous Velocity: On a position-time graph, this is found by calculating the gradient of the tangent line touching the curve at that specific point. Mathematically, it is the limit of the average velocity as the time interval shrinks to zero ($\Delta t \to 0$).
  • Instantaneous Speed: This is simply the magnitude (absolute value) of the instantaneous velocity. It tells you how fast the object is moving at that exact moment without considering its direction.
$$v = \lim_{\Delta t \to 0} \dfrac{\Delta x}{\Delta t}$$
t x Tangent at P P

Example 1

t (s) x (m) 0 30 -10 2 5 10

Problem: A car drives into an alley, finds it blocked due to construction, and has to reverse out in a straight line. Its position ($x$) versus time ($t$) relationship is shown in the graph above. Determine:

  1. The instantaneous velocity at $t = 1\text{ s}$
  2. The average velocity from $t = 0\text{ s}$ to $t = 10\text{ s}$
  3. The average speed from $t = 0\text{ s}$ to $t = 10\text{ s}$

Solution:

  • a) Instantaneous velocity at 1 s:
    The instantaneous velocity is the gradient of the line at $t = 1\text{ s}$. Since the line is straight from $0\text{ s}$ to $2\text{ s}$, we calculate the gradient of this entire segment:
    $$v = \dfrac{\Delta x}{\Delta t} = \dfrac{30 - 0}{2 - 0} = 15\text{ m/s}$$
    Instantaneous velocity = 15 m/s (forward)
  • b) Average velocity from 0 to 10 s:
    Average velocity is the total displacement divided by total time. The total displacement $\Delta x = x_{\text{final}} - x_{\text{initial}} = -10 - 0 = -10\text{ m}$.
    $$\text{Average velocity} = \dfrac{\Delta x}{\Delta t} = \dfrac{-10}{10} = -1\text{ m/s}$$
    Average velocity = -1 m/s (backward)
  • c) Average speed from 0 to 10 s:
    Average speed is the total path length divided by the total time. The car moves forward $30\text{ m}$ (from $t=0$ to $2\text{ s}$) and then moves backward $40\text{ m}$ (from $x=30$ to $x=-10$). Total distance = $30 + 40 = 70\text{ m}$.
    $$\text{Average speed} = \dfrac{l}{\Delta t} = \dfrac{70}{10} = 7\text{ m/s}$$
    Average speed = 7 m/s

Velocity-Time Graphs

A velocity-time graph is highly informative because both its gradient and its enclosed area correspond to kinematic quantities.

Gradient = Acceleration

Since the y-axis is velocity ($v$) and the x-axis is time ($t$):

$$\text{Gradient} = \dfrac{\Delta v}{\Delta t} = a$$

Area = Displacement

Because displacement is velocity multiplied by time ($\Delta x = \int v \, dt$), the space bounded between the graph line and the time axis is the displacement.

  • If the enclosed area is located above the time axis, the displacement is positive.
  • If the enclosed area is located below the time axis, the displacement is negative.
$$\text{Area under graph} = \Delta x$$

Summary of Velocity-Time Graph Principles

Based on graphical calculus principles, we analyze $v-t$ layouts using three essential parameters:

  • Average Acceleration ($\overline{a}$): Bounded by any two points on a $v-t$ curve, the average acceleration is the slope of the straight secant line connecting them:
    $$\overline{a} = \text{Secant Slope} = \dfrac{\Delta v}{\Delta t}$$
  • Instantaneous Acceleration ($a$): Bounded at any exact moment in time, the instantaneous acceleration is the slope of the localized tangent line touching that point on the curve:
    $$a = \text{Tangent Slope} = \dfrac{dv}{dt}$$
  • Displacement ($\Delta x$): The net change in position is equivalent to the definite integrated geometric area trapped between the function line and the zero horizontal time axis:
    $$\Delta x = \text{Enclosed Area} = \int v \, dt$$

Visualizing Constant Slopes (Uniform Motion): If a line on a $v-t$ graph has a fixed, unchanging slope, the object's acceleration is constant at every moment, characterizing uniform acceleration motion.

t (s) v (m/s) 0 1 2 3 4 1 2 3 4 5 Line I (a = +0.25 m/s²) Line II (a = -0.5 m/s²) Δt Δv Δt' Δv'

Example 2

Velocity-Time (v-t) Profile

t (s) v (m/s) 0 10 -5 4 6 8 10

Corresponding Acceleration-Time (a-t) Profile

t (s) a (m/s²) 0 2.5 -2.5 -5 4 6 8 10 a₁ a₂ a₃ a₄

Problem: An elevator in a department store moves up to the top floor (moving fast then slow), stops for 2 seconds, and then begins to descend. Calculate:

  1. The total displacement of the elevator over the first 10 seconds
  2. The average velocity of the elevator over the first 10 seconds
  3. The acceleration of the elevator during the first 4 seconds
  4. The highest height reached by the elevator during this 10-second period

Solution:

  • a) Total Displacement:
    The displacement is the area enclosed between the velocity-time graph and the time axis. The area above the time axis is positive displacement, and the area below is negative displacement.
    $$\text{Area}_{\text{above}} = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 6\text{ s} \times 10\text{ m/s} = 30\text{ m}$$
    $$\text{Area}_{\text{below}} = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times (10\text{ s} - 8\text{ s}) \times (-5\text{ m/s}) = -5\text{ m}$$
    Total Displacement = 30 m - 5 m = 25 m
  • b) Average Velocity:
    Average velocity is the total displacement divided by the total time interval.
    $$\overline{v} = \dfrac{\Delta s}{\Delta t} = \dfrac{25\text{ m}}{10\text{ s}}$$
    Average Velocity = 2.5 m/s
  • c) Acceleration (0 to 4 s):
    The acceleration is the gradient of the velocity-time graph during this specific interval.
    $$a_1 = \dfrac{\Delta v}{\Delta t} = \dfrac{10\text{ m/s} - 0\text{ m/s}}{4\text{ s} - 0\text{ s}} = \dfrac{10}{4}$$
    Acceleration = 2.5 m/s²
  • d) Highest Height Reached:
    The elevator reaches its highest height exactly when it finishes moving upward (when positive velocity reaches 0). This happens at $t = 6\text{ s}$ and it remains at this height until $t = 8\text{ s}$. This height corresponds to the total positive area under the graph.
    Maximum Height = $\text{Area}_{\text{above}}$ = 30 m

Interpreting the lines:

  • Horizontal Line: The velocity stays the same. The object moves with constant velocity ($a = 0$).
  • Straight Sloped Line: The velocity increases by the same amount each second. The object undergoes uniform acceleration.
  • Curved Line: The velocity increases by a larger amount each second. The object undergoes non-uniform acceleration.