Often quantities found in equations can be related to each other in a straightforward way. We can write simple relationships between these quantities to make solving problems a little easier.
For example, take the equation for average velocity,
\(V_av = \frac {d}{t}\)
The average velocity is calculated as the distance travelled, d divided by the time taken, t.
If the time taken is kept fixed, but I travel at a higher velocity, I will travel further (d is greater).
Notice that for a constant time interval, if I double V (travel twice as fast), the distance travelled will also double. We can say that the velocity is directly proportional to the distance travelled.
General Rules:
When we say variable quantities x and y are directly proportional, we mean that as x and y change, the ratio x/y is constant. To say that two quantities are proportional is to say that they are directly proportional. When we say variable quantities x and y are inversely proportional, we mean that as x and y change, the ratio xy is constant.
Relationships of direct and inverse proportion are common in physics.
Examples
Test your understanding with these true/false questions
When two quantities are directly proportional, the two quantities are related by a constant of proportionality.
For example - If you are paid for working at a regular rate R in dollars per day, the money M you earn is directly proportional to the time, t you work.
Sometimes the constant of proportionality can be ignored in proportion problems. Because the amount you earn in 8 days is 8/5 times what you earn in 5 days, this amount is
m8 days = 8days($400/5days) = $640
We will look at this idea more in the following section.
In the simple equation, x = zy
If z is a constant, x is directly proportional to y.
This relationship can be written as
\(x \propto y\)
If two quantities are directly proportional then calculating x/y will always comes out as the same value - provided there are no other non-constant variables in the equation.
This means we can write another relationship
x0/y0 = x1/y1
The following worked example will demonstrate how this can be useful.
If you stretch a spring a distance d, it exerts a restoring force F on your hand. The force depends on how much you stretch the spring and this relationship is described by the equation,
F=-kd
Question: If the spring is initially stretched by 2cm from its relaxed position, I measure a restoring force of 10N. Calculate what the restoring force would be if I instead stretched the spring by 6cm from its rest position.
Solution: To solve this problem you could first solve for k using d = 2cm and F = 10N, and then use the value to find the value of F for d = 6cm, but we will do it a different way.
Since
F = -kd and –k is a constant, \(F \propto d\)
This means I can use the rule, F/d is a constant
We can write a new expression
F0/d0 = F1/d1
Where F0 and d0 represent the numerical values for the initial stretch and F1 and d1 represent the values for the second stretch.
Rearranging this expression gives,
F1 = F0d1/d0
Now you can solve for F1
F1 = 10N(6cm)/2cm = 30N
Notice that using this method we did not calculate a value for k, it was not needed to solve the problem.
Accelerating motion can be described by the equation F = ma where an acceleration, a results from an applied force, F on an object of mass m.
The potential energy, U stored in a stretched spring is given by the equation,
U = ½kd2
where k is the spring constant.
CJRY6/yNMbqw7terMPmh4OHLvsILTeo9QWE3v3oDVdFumRf3BZ/y5UVykcfruJH9IO0t+X6TT0u4LYvt6ENj44X1mLzrTxtePCE+jDYGVcgeKMUdiNlWyM8QlpI5+cJhZly1oljDvrBde/Eauba3lhM6z2Dx3qt2e9M3tWXYIrwl/PDXzAzc0DmIRMzEBfmDWNrE85so40+hcopX33jQRyN5HGzh6ByADhQ/dyuKEBdHtWemqptx9o2GjPCdCjJsPlTajvgyXoiAzx1sU+FhLqdVcOYvvrdAZDUowiFCnERfondVURAs0LDZ8UOD6nTkDcz7iW+zLKftGEe+KCSdXzzMa672AuyWzQKDjCEKA+6/Ye5b5mg+AzrvndUARc+Xdy3YWEreM3O7m/2l8sKJewn1la0ualpQ2eVLcOLyFngEwJYl