Teacher tools for this standard
Lesson Plan · Guided Notes · Exit Ticket · Re-teach · Homework
Teacher tools for this standard
Lesson Plan · Guided Notes · Exit Ticket · Re-teach · Homework
- Lesson Plan →Objectives, pacing and practice, built from this lesson's brief.
- Guided Notes →One page your students fill in and keep.
- Exit Ticket →Three items at the end of class. No student accounts.
- Re-teach →After an exit ticket: who missed what, and what to do tomorrow.
- Homework →Assign practice; it grades itself.
"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"
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"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"
What you'll learn
- 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
Slides
Step through the lesson, or watch it as a narrated video
Slides
In development
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