On a velocity-time graph, what does the slope of the line at a moment give?
Here is the car from the first kinematic equation, drawn as a velocity-time graph. It pulled away from the light at 3 m/s and accelerated at 2 m/s² for 5 s, reaching 13 m/s.
At t = 0 s the car already moves at 3 m/s; five seconds later it moves at 13 m/s. Every part of v = v₀ + at is somewhere in this picture.
The line starts at v₀: at t = 0 s, the car is already moving at 3 m/s.
The line's slope is a. You've already seen that the slope of a velocity-time graph is the acceleration — here, 2 m/s of extra velocity for every second across.
And at t = 5 s, the line has reached v: 13 m/s. The graph is v = v₀ + at, drawn.
Two students read the same velocity-time graph. Student P says the area between the line and the time axis gives the displacement. Student Q says the area gives the distance traveled, so it can never count as negative. Who is right?
One more thing lives in this picture. You've already seen that the area under a velocity-time graph is the displacement.
So the same drawing holds the whole story: the line is v = v₀ + at, and the area beneath it is Δx — how far the car went.
One picture, one equation: where the line starts, how it climbs, where it gets to — and, underneath it, how far the car went.
A velocity-time graph shows a straight rising line. In v = v₀ + at, which quantity is the line's slope?
A velocity-time graph shows a straight rising line. Where on the graph is v₀?
On a velocity-time graph with a straight rising line, the region between the line and the time axis is shaded. Which quantity does that area give?