Lesson Plan

Identify the independent variable (time, horizontal axis) and dependent variable (position, vertical axis) on a motion graph

Explain that the slope of a position vs. time graph is the velocity

Read the initial position d_0 as the y-intercept and connect the graph to d = d_0 + v t

Calculate average velocity from a straight-line position–time graph using two points

Find instantaneous velocity on a curved position–time graph as the slope of the tangent line

2.3

"When two physical quantities are plotted against each other, the horizontal axis is usually considered the independent variable, and the vertical axis is the dependent variable." "As shown, a straight-line graph has the general form $y = mx + b$. Here m is the slope, defined as the rise divided by the run ... The letter b is the y-intercept which is the point at which the line crosses the vertical, y-axis." "The rise is the change in position, (i.e., displacement) and the run is the change in time. ... Therefore, the slope in a d versus t graph, is the average velocity." "$\vec{d} = \vec{d}_0 + \vec{v}t.$" "Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525 m). ... $\vec{v} = \frac{\Delta\vec{d}}{\Delta t} = \frac{2000 \text{ m} - 525 \text{ m}}{6.4 \text{ s} - 0.50 \text{ s}} = 250 \text{ m/s}$" "The slope at any point on a position-versus-time graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line." "These correspond to a position of 1,300 m at time 19 s and a position of 3120 m at time 32 s. ... $\text{slope} = v_Q = \frac{\Delta d_Q}{\Delta t_Q} = \frac{(3120 - 1300) \text{ m}}{(32 - 19) \text{ s}} = \frac{1820 \text{ m}}{13 \text{ s}} = 140 \text{ m/s}$" "tangent: a line that touches another at exactly one point"