Lesson Plan

Derive a velocity vs. time graph from a position vs. time graph

Explain that the slope of a velocity–time graph is the acceleration and the area under it is the displacement

Read instantaneous velocity directly off a velocity–time graph, and find instantaneous acceleration from the tangent on a curved graph

Compute displacement as the area under a straight velocity–time graph by decomposing it into rectangles and triangles

Compute acceleration as slope and average velocity as net displacement divided by total time from a velocity–time graph

2.4

"Velocity is the rate of change of displacement. Acceleration is the rate of change of velocity; we will discuss acceleration more in another chapter." "First, we can derive a v versus t graph from a d versus t graph. Second, if we have a straight-line position–time graph that is positively or negatively sloped, it will yield a horizontal velocity graph." "The area under a velocity curve represents the displacement. ... the slope equals the acceleration, a. And in this graph, the y-intercept is v₀. Thus, $\vec{v} = \vec{v}_0 + \vec{a}t$." "In this case, the area is made up of a rectangle between 0 and 20 m/s stretching to 30 s. The area of a rectangle is length × width. Therefore, the area of this piece is 600 m. ... Above that is a triangle whose base is 30 s and height is 140 m/s. ... the area of this piece, therefore, is 2,100 m. ... Add them together to get a net displacement of 2,700 m." "$\vec{a} = \frac{\Delta\vec{v}}{\Delta t} = \frac{100 \text{ m/s}}{20 \text{ s}} = 5 \text{ m/s}^2$" "Divide 2,700 m / 30 s = 90 m/s." "Add them together to get a net displacement of 16,325 m. (b) Using the tangent line given, we find that the slope is 1 $\text{m/s}^2$. (c) The instantaneous velocity at t = 30 s, is 240 m/s." "You can have negative position, velocity, and acceleration on a graph that describes the way the object is moving. You should never see a graph with negative time on an axis." "The slope of a velocity vs. time graph is the acceleration. The area under a velocity vs. time curve is the displacement."