The concept of acceleration– in everyday experience
By Tom Brown
Tue, 20 May 2014
When on a short break to the farm recently my brother asked me some questions by our little fire under the Southern Sky. It gave me the idea for this essay and it was written for him.
It could be classified as popular science. It is rudimentary (classical) mechanics on the level of secondary school science. I've tried to write it in such a way that one may grasp the ideas without the formulas. Still, it would probably take a bit of effort. It turned out longer than what I'd planned originally but the text is divided into more digestible parts that can be read independently, and there are three shorter essays. One can skip and come back later, or as you like. If someone is really interested in the material it would go smoothly enough.
The essay should be understandable to everyone and there are no prerequisites apart from curiosity and imagination.
To begin with we need clarify a few things and explain some ideas. The concept of velocity (or speed) is very well known and is well understood in everyday life, together then obviously with distance travelled and time taken. Momentary speed is the speed at that point or particular moment, whether the speed is increasing or decreasing or constant. When your speed is changing, say you're pulling away and you're going faster and faster. The speedometer reading will be steadily increasing from zero and the speed at any given moment in time is the reading at that moment, your momentary, or instantaneous speed.
As an illustration: On the open road if you cover 500km's in 4 hours you would have been travelling at an average speed of 500/4 = 125km/h. Say you read the same speed on your speedometer all the way- it would be constant at 125km/h. In this case the momentary speed at any given time is the same as the average.
However the concept acceleration is in general not so well understood nor is it correctly used. In everyday language it could be defined as the rate of change of velocity. But “average acceleration” is easy to define and to understand. It is the change of velocity divided by the change in time in a given interval.
As an example suppose now I am burning a dragster from standstill. If my speed increases to (and by) 40m/s in the first two seconds from zero, the average acceleration would be 40/2 = 20m/s² or just over 2 “g's”.
Average acceleration is all that is needed here since we will only consider problems where acceleration is constant i.e. doesn't change with time, in which case it is always equal to the average. Please note that technically the concept of velocity is not quite the same as that of speed but for current purposes we may take them as the same. Similarly there is a difference between the terms displacement and distance which is not important here.
The study of motion from a purely spatial viewpoint is called kinematics. It does not involve the reasons, or physical causes and effects but only the movement itself.
Variables and Units
The unknowns are denoted by s, t, v, u, a (g). Distance, time, velocity, initial velocity and the acceleration. For instance when the object is free-falling a would be gravitational acceleration g. It will be assumed that all motion has constant acceleration and that it is in a straight line. In the equations I take all motion from standstill, i.e. v (= u) = 0 when time t = 0. It simplifies matters and it is easy to adapt to v = u ≠ 0 at t = 0.
Metric units are to be used. Distance is measured in metres ( m ), time in seconds ( s ), velocity in metres per second ( m/s ) and acceleration ( m/s² ).
The equations of simple linear motion
These have only to do with time and space and don't involve force nor mass nor anything else, and with acceleration as given.
v = at
s = ½ at²
v² = 2as
s = ½ at²
v² = 2as
Of course if a = 0, when there is no acceleration, the velocity v = u stays the same as the initial velocity. It is constant and gives the familiar s = ut = vt.
In the three equations there are four unknowns: s, t, and v are variables, and the constant a (assuming u = 0).
Already at this stage some useful calculations can be made. If only one of the variables and of course the fixed a are known, the other two may be found. For instance we may find the position s, the distance from the starting point, and the velocity of our object at any given time.
A few practical calculations
In drag racing the driver experiences extremely strong forces due to large acceleration from pull-away as a result of very powerful rocket or nitrogen-oxide engines. They race on a straight track of fixed length and the time is measured. Faster vehicles cover the distance quicker that is to say in less time.
The distance is known and the measured stopwatch time so that one can use the equation s = ½ at² to solve for a. It works out a = 2s/t² with s being the length of the straight in meters and the timed t is in seconds. Acceleration is then in m/s². For the g's one can just divide a by g = 9.8m/s². For example 5g's will be 5 x 10 = 50m/s² with g then rounded off to 10m/s², 9.8m/s² is a bit more accurate.
As another example a sport-car's performance is often given in seconds for 0 – 100km/h. Then the acceleration is easy to calculate. The first formula v = at gives a = v/t where v is 100km/h and t is the clocked time. One must just be very careful with your units. In these calculations a is the average, but in reality the acceleration most probably is not constant. Even said, it is so that the peak, the maximum value over a stretch is always more than the average.
When acceleration is due to gravity, a is constant relatively near the earth surface and is the constant g with a = g ≈ 10m/s². In what follows I will be referring to the three equations given above.
A stone falls into a well and we time it. We may now calculate the well's depth and the stone's end velocity. If we let it fall and we hear it splash after 3 seconds, the second equation gives the depth, the distance fallen as s = ½ at² = ½ gt² = ½ x 10 x 3² = 45 meters and from the first, v = gt = 10 x 3 = 30m/s.
Suppose I drop a brick from a tall building and I know the height of the building. I may solve the second equation for time t = √(2s/g) with s the height and the first equation then gives its end velocity. If a stone falls from a cliff of known height it is exactly the same. You can calculate the time it takes to fall as well as the impact velocity.
If the impact velocity is known (say from forensic evidence) we may calculate time and height. On solving the first equation for time, t = v/g the second one then easily gives the height, distance fallen. This is essentially the same as our sport-car example. You can do the same for say a rocket or a fighter jet (not diving or swerving because centrifugal forces will complicate things considerably!)
The examples were objects that fall freely, vertically under the influence of gravity only.
Thus I may measure distance indirectly, by measuring time. It works the same with the more well known method of finding the distance to lightning. The speed of sound in ordinary circumstances is about 332m/s so that if we time from the lightning flash to the thunder, if I hear it after 3s it means it was about 1km away.
In all fairness we should take the speed of sound into account when finding the depth of the well too although wells seldom are that deep. However in principle it is not hard to do. You would write two equations for the distance of fall. One is the drop and one the sound, back. The result would be a straightforward quadratic equation, solved easily by the familiar formula.
A falling rock and a lightning bolt are like a car crash. When you see it coming it has already happened.