Uniform acceleration occurs when the speed of an object changes at a constant rate. The acceleration is the same over time. By relating acceleration to other variables such as speed, time and distance we are able to manipulate data in many ways.
We have a small set of very powerful formulae which can be combined to obtain results despite unknown variables. These formulae have been deduced in the Visualizing Motion section.
a = ( Vf - Vi ) / t
This is the formula for the average acceleration, which is our actual acceleration when dealing with uniform acceleration. Average acceleration equals to the difference in speed over time. When we have two speeds we can calculate the acceleration during the time interval, t, by subtracting the initial speed from the final speed ( we get the change in speed ) and then dividing by the time to get the acceleration.
Acceleration = increase in speed / time
d = 1/2 ( Vf + Vi ) × t
This formula is a favourite of mine. It stems from analyzing the graph of speed V.S. time of an object's motion. Since speed = distance / time we can calculate the distance ( or displacement ) an object by measuring the area under the line which represents speed. The derivation of this formula is quite simple.
You now know two formulae that relate the initial speed, Vi, the
final speed, Vf, the distance, d, the acceleration, a
and the time, t.
When you are solving a problem where you need to find one of these variables but you are lacking another one, you should combine the two formulae to eliminate the unknown variable. Here is a list of the combinations you can use when not given a variable:
Not Given Formula to eliminate that variable
( not needed )
d a = ( Vf - Vi ) / t
a d = 1/2 ( Vf + Vi ) × t
Vf 2d / t = at + 2Vi
Vi 2d / t = -at + 2Vf
t 2da = Vf² - Vi²
These different combinations of the same two formulae are tedious to memorize and I suggest that you remember the original formulae only. When you need to combine the formulae do so algebraically, instead of memorizing each combination.
Lets see just how useful is the ability to eliminate a variable.
A car was travelling at a speed of 70km/h, the driver saw a rabbit on the road and slammed on the breaks. After 6.0 seconds the car came to a halt, how far did the car travel from the point where the brakes were first pressed to the point where the car stopped?
Here our knowledge of uniform acceleration is very useful. We are not given the acceleration of the car as its stopping ( we assume the rate of acceleration is uniform ) and we need to find the distance.
We are given:
Vi = 70km/h = 19.4 m/s
Vf = 0km/h
t = 6s
d = ?
Our formula for distance is d = 1/2 ( Vf + Vi ) × t
d = 1/2 ( 19.4 + 0 ) × 6
d = 58.332
The car stopped after 58 metres.
This car has very bad brakes, as 58 metres is a very long distance to come to a halt.
Lets do an example where we can combine the different formulae to get results.
Bill jogs at 6.0km/h, he then decides to accelerate into a light run.
Bill accelerates at 0.030km/s² as he runs through a distance of 40m
What is Bill's final speed?
First we convert our given information into a uniform set of magnitudes, metres and seconds. We are given:
Vi = 6.0km/h = 1.6m/s
Vf = ?
d = 40m
a = 0.003km/s² = 0.30m/s²
Lets rearrange our uniform acceleration equations to eliminate the time, which we are not given.
2da = Vf² - Vi²
Vf² = 2da + Vi²
Vf² = 2(40)(0.3) + 1.6²
Vf² = 26.6
Vf = 5.2m/s
Vf = 18.5km/h
After accelerating Bill ends up running at 18 km/h.
When an object is accelerating at a constant rate its motion can be modelled by two simple equations, a = ( Vf - Vi ) / t and d = 1/2 ( Vf + Vi ) × t . Using these equations gives you the ability to discover information about the motion while lacking a variable. The equations can be rearranged and substituted into each other to compensate for the lack of the distance, initial speed, final speed, acceleration or the time variable.
It is complicated to memorize every single arrangement of the two equations and I suggest that you practice creating new combinations out of the original equations. Algebraically.