# An approximate solution for constant acceleration

Teaching Guidance for 14-16

The following is suggested as a possible first example, for students, of numerical integration.

The phenomenon of a falling ball, accelerating at close to 10 m s^{-2}, can quickly be given a graphical solution. Knowing that the distance fallen is given by *s* = ½*a**t*^{ 2} also makes it possible to check the graphical technique.

Suppose the ball starts at rest, at time t = 0. Over a short time, say 0.1 second, around *t* = 0 the velocity is pretty well zero, the distance travelled is also nearly zero, and the first bit of the graph must be flat, like the segment AB in the figure above.

But in the following interval of another 0.1 second, around the 'time-of-day' t = 0.1 second the average velocity will equal that at that time. If the acceleration is 10 metres per second each second, the velocity is 1.0 metre per second. So the next segment of the graph, BC, in the same figure, rises or slopes up at 1.0 metre per second, rising in all 0.1 metre in an interval of 0.1 second.

The next 0.1 second interval centres around the time 0.2 second from the start, when the velocity will be averaging 10 x 0.2 = 2 metres per second. So the next section of the graph slopes up twice as steeply, rising 0.2 metre over 0.1 s econd. And so it goes on. The rule is easy: around each time *t* draw a section of line for velocity 10 *t* which must rise an extra distance (0.1) 10 *t* metre.

Note that for successive equal time intervals Δ*t*, the velocity rises by the *same amount each time*. The acceleration is constant.

Such a graph is shown above (figure 2). The circles mark points calculated from *s* = ½*a**t*^{ 2}, for comparison. It predicts, for instance, that a heavy ball will fall 1 metre in 0.45 second. This could be tested using a scaler to time the fall, if your class feels the need. Or a ball could simply be dropped, and the class be asked to estimate the time.
The equation *s* = ½*a**t*^{ 2}} and the kinked curve

are both solutions

of the equation:

acceleration = 10 m s^{-2}
or, better, of
= *a* where *a* = 10 m s^{-2}.

The graph is an QuoteThis{approximate solution: it is near to the exact solution, and although where it is wrong it always makes the distance come out a little too small, the graph does not drift off course (that is, the errors do not accumulate).
The smaller the interval Δ*t*, the better the approximation. When solving real problems in this style, engineers and physicists devote much attention to choosing Δ*t* small enough to be just sufficiently accurate, but not so small as to make the job unnecessarily tedious. Students are likely to have more confidence in the idea if the problem of ‘sufficiently good’ approximations is taken seriously.

(Can any standard school-level apparatus, in fact, detect an error of the size there appears to be between the graph and results found from *s* = ½*a**t*^{ 2}?)

## How constant acceleration is represented in drawing the graph

See figure 3 above. In drawing the graph, each new section was drawn at a steeper slope; that is, a larger velocity. Because the acceleration was constant, the slope increased by equal amounts in each step.

See figure 4. At each moment the average velocity near that moment was worked out, using *a* = 10 m s^{-2}. A line like AB in the above figure at the correct slope was put in, going an extra distance *v*Δ*t*. At the next moment, *v* was larger, say *v* + Δ*v*, where *v* is the extra velocity gained in an interval Δ*t*. A line such as BC was drawn, at the new, larger slope, going a larger extra distance (*v* + Δ*v*) Δ*t*.

If, as in figure 5, AB is run on at the same slope, to D, then DF = *v*Δ*t* is the extra distance that would have been covered if the velocity had *not* increased. The extra extra

bit CD = Δ*v*Δ*t* is the 'extra extra distance’ gone because the velocity *did* increase, by Δ*v*.

So CD = Δ*v*Δ*t* = *a*(Δ*t*)(Δ*t*) = *a**(Δt)*^{ 2}

## A rule for drawing graphs of acceleration

Looking at figure 6, there is a simple rule for drawing acceleration graphs. Take the graph AB as found at the last interval, and run on AB straight to D, as if there were no acceleration. Then add an extra extra distance

DC, where DC = *a**(Δt)*^{ 2}. Then BC is the next bit of graph. If the acceleration should vary, the extra extra distances

like DC will vary; just work them out from the basic recipe.

## The rule with a Δ notation

For some students, the argument in terms of acceleration will be more than enough. But others may like to see how the calculus notation is useful. Δ is used to mean small change in . . .

.

Over AB, in figure 7, the velocity *v* is the slope of AB, that is Δ*s*Δ*t*. Then the velocity rises, and the acceleration is

Over the timeΔ*t*, the distance *s* changes by more than it would have changed if there had been no acceleration, and Δ*s* in the second interval is larger than Δ*s* in the first by the ‘extra extra distance’ Δ(Δ*s*). Further, the acceleration is Δ(Δ*s*)Δ*t*^{ 2}, equal to the rate of change of velocity around B.

So the rule remains the same: add on an 'extra extra distance' over and above that which would come from taking the graph straight on. The size of the 'extra extra distance' should be:

Δ(Δ*s*) or Δ^{2}*s* = (acceleration)Δ*t*^{ 2}.

Δ*s*Δ*t* approximates to the velocity *v*, or d*s*d*t*.

Δ^{2}*s*Δ*t*^{ 2} approximates to the acceleration *a*, or *d*^{ 2}*s**d**t*^{ 2}.