Welcome to practical physicsPracticle physics - practical activities designed for use in the classroom with 11 to 19 year olds
 

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 vabove 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 speed

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. 

Constant acceleration

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.

The area represents 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)

The area of any triangle is 1/2 (height) x (base). So distance s is 1/2 (height, v) x (base, ts = 1/2vt
 
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). 

Suppose the object does not start from rest

The rectangle and the triangle

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 
= 1/2vt + 1/2ut 
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 + 1/2at2 
 
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 = 1/2at2 is no longer safe for calculating s

What part of ut = 1/2at2 is no longer safe for calculating s?