# Using speed-time graphs to find an equation

Imagine a graph plotted with SPEED on the vertical axis against TIME on the horizontal axis.

**Constant speed**

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*.

**Constant acceleration**

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, *v*_{1}. 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 *v*_{1} 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 1/2 (height) x (base). So distance *s* is 1/2 (height, *v*) x (base, *t*) *s* = 1/2*vt*.

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 1/2(*v - u)t*.

Then *s* = *ut* + 1/2 (*v-u)t*

= *ut* + 1/2 *vt* - 1/2 *ut*

= 1/2*vt* + 1/2*ut*

*s* = (*v* + *u* / 2)*t*

Alternatively, since *v - u* = at

*s* = *ut* + 1/2 (*v - u*)*t* can be expressed as

*s* = *ut* + 1/2 (*at*)*t*

= *ut* + 1/2*at*^{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* = 1/2*at*^{2} is no longer safe for calculating *s*?