Velocity vs. Time Graphs

"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 *v*₀. 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."