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Some more advanced ideas in maths.
Calculate velocity from distance/time graph
Calculate velocity from distance/time graph
If we plot a graph of distance (s) versus time (t) we can see how an object moves.
In graph 1 we see that the distance increases linearly with time (blue line), at a velocity of 1 metre per second, because the distance increases by 1 metre for every second.
If the object moved at a higher velocity the line would have a steeper slope - the slope of the line tells us the velocity.
If the line is curved (e.g. distance s = t2), like the red curve in graph 2, we know that the velocity is changing. In this case it is increasing.
If we could judge the slope of the line at a certain time, we would know the velocity at that time.
Let's say we want to estimate the slope at 3 secs. If we add 1sec (as shown by the black triangle on the graph), and work out the change in distance it will give us the slope (velocity).
But you can see that the slope of the triangle does not correctly match the slope of the curve at 3secs.
Adding a smaller time (say 0.1secs or 0.01secs) makes the estimate more accurate (as shown in the table), and the slope tends towards 6.
This method is the basis of differention.
We can do this mathematically as follows:
In this case the equation of the line is s = t2
So s+Δs = (t+Δt)2 = t2 + 2tΔt +Δt2, where Δs and Δt are the small changes in distance (s) and time (t).
As Δt goes to a very small value, this becomes s+Δs = t2 + 2tΔt
After subtracting s = t2 , we get Δs = 2tΔt, so
Δs/Δt = 2t.
So we know that the slope (velocity) is 2t, for any t.
At 3secs the velocity is 6metres/sec, as we estimated earlier.
Differentiation tells us the rate of change of one term versus another.
In differentiation we write Δs/Δt as ds/dt.
We also find that the general differentiation rule for
y = xn is dy/dx = nx(n-1) , where y and x can be any parameters, and n is any power.
A straight line means a constant velocity. The slope of the line tells us the velocity.
An upward curved line means the velocity is increasing with time (in this case s = t 2).
To calculate the velocity at 3secs, we can add a small amount of time and calculate how much that increases the distance. The shorter the added time the more accurate the velocity becomes:
Can we calculate distance from velocity?
Can we calculate distance from velocity?
In some cases we might want to do the opposite calculation - i.e. work out the distance from the velocity/time graph.
Let's plot a velocity vs time graph, see graph 3, for a constant velocity of 1m/s. (This is the same situation that we studied in graph 1 earlier).
We also know that distance is velocity x time.
So if the speed is constant this is equivalent to the area under the line, see the blue area on the graph.
The area of the shaded region under the line (3x1=3) shows that a distance of 3 metres was travelled after 3 secs at a constant speed of 1m/s.
As time increases the area under the curve increases, giving a larger distance travelled (4m at 4 secs, 5m at 5secs, etc).
So we get back to the distance vs time relationship we saw in the graph 1 earlier, proving that the 'area under the curve' method works.
If the velocity is changing, e.g. sloped or curved, then we can still apply the same 'area under curve' rule.
In graph 4 we have plotted another case: where velocity is 2 x time, just as we saw in graph 2 earlier.
We can use the triangle area method (½ x base x height) to test the 'area under curve' method again.
At 1sec this is ½ x 1 x 2 = 1, at 2secs it is ½ x 2 x 4 = 4, at 3secs it is ½ x 3 x 6 = 9, etc.
This matches the distances we saw in graph 2, showing that the area method works again.
If the line is curved, we would need to break the area into small vertical sections, as in graph 5, then add them all together to find the total area.
We can write this as: v1.Δt + v2.Δt + v3.Δt... , where Δt is the width of the strips.
Written more formally: ∑vn.Δt , where ∑ means 'sum of'.
The narrower the sections, the more accurate the area calculation.
This method is the basis of Integration.
It is basically calculating the area under the curve.
We can see from the simple examples above that the integration is the reverse of differentiation.
Or the 'integral' is the 'anti-derivative'.
We would write the integral as: ∫v.dt
Because the area under the curve depends on where we are measuring from and to, we typically need to specify the limits of the integration.
We know that if y = xn , then dy/dx = nx(n-1)
So then ∫nx(n-1) .dx = xn + C
Or written another way ∫xn .dx = x(n+1)/(n+1) + C
We add a constant C when we don't have any limits of integration.
When we differentiate a constant it disappears, so we add it to the integral answer (because we can't be sure that there wasn't a constant there before differentiation), so this is the most correct answer.
We may find that C = 0, but we must first check that.
The value of the constant C can be found if you know the y value for a given x, e.g. at x = 0.
The area under the curve tells us the distance travelled:
If we have a sloped or curved line we can still use the 'area under the curve' rule:
If the line is curved then we need to break the area into small blocks:
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