## 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.

## 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 ½ (height) x (base).

So distance *s* is ½(height, *v*) x (base, *t*) *s* = ½*v**t*.

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 *u**t*, the triangle's is ½(*v*-*u*)*t*.

Then *s* = *u**t* + ½(*v*-*u*)*t*

- =
*u**t*+ ½*v**t*-½*u**t* - = ½
*v**t*+ ½*u**t* - = (
*v*+*u*2)*t*

Alternatively, since *v*-*u* = *a**t*

*s*=*u**t*+ ½(*v*-*u*)*t*can be expressed as- =
*u**t*+ ½(*a**t*)*t* - =
*u**t*+ ½*a**t*^{ 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 *u**t* = ½*a**t*^{ 2} is no longer safe for calculating *s*?