Using speed-time graphs to find an equation
Teaching Guidance for 14-16
Imagine a graph plotted with SPEED on the vertical axis against TIME on the horizontal axis.
For an object moving along with constant speed v, the graph is just a horizontal line at height v above the axis. You already know that s, the distance travelled, is speed multiplied by time, vt; but on your graph v x t is the AREA of the shaded block of height v and length t.
Sketch a graph for an object starting from rest and moving faster and faster with constant acceleration. The line must slant upwards as v increases. And if the acceleration is constant the line must be a
straight slanting line.
Take a tiny period of time from T to T' on the time axis when the speed was, say, v1 . Look at the pillar that sits on that and runs up to the slanting graph line (Graph III). The area of that pillar is its height v1 multiplied by the short time TT'. That area is the distance travelled in that short time.
How big is the distance travelled in the
whole time, t, from rest to final v? It is the area of all the pillars from start to finish. That is the area of the triangle (in Graph IV) of height final v and base t, the total time.
The area of any triangle is ½ (height) x (base).
So distance s is ½(height, v) x (base, t) s = ½vt.
Suppose the object does not start from rest when the clock starts at 0 but is already moving with speed u. It accelerates to speed v in time t. Then the graph is like graph V below; and the distance travelled is given by the shaded area. That is made up of two patches, a rectangle and a triangle (Graph VI).
The rectangle's area is ut, the triangle's is ½(v-u)t.
Then s = ut + ½(v-u)t
- = ut + ½vt-½ut
- = ½vt + ½ut
- = (v + u2)t
Alternatively, since v-u = at
- s = ut + ½(v-u)t can be expressed as
- = ut + ½(at)t
- = ut + ½at 2
These formulae are only true for constant acceleration. Look at Graph VII. Is the acceleration constant? Which part of the area for s is different now?
What part of ut = ½at 2 is no longer safe for calculating s?