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Next page : Acceleration - calculating the time
We will continue with the rounded value of the acceleration of gravity on the Earth as a=g=10 m/s2.
We can find the height at which an thrown object is at by using a v/t-diagram, as we saw on the precious page, or by using the formula we found on the previous pages. To recapitulate, how far has a dropped object fallen after say 3.0 seconds?
If we choose up to be in the positive direction, then a v/t-diagram would look something like this:
This could, for example, correspond to a fall from 0 m and 45 m down, or from a hight of 45 m down to the ground.
We could of course us the formula
This wound give us s= –10·32/2= –10·9/2= –90/2= –45 m as before.
Time from the displacement
Yet again, let us say we start at 0 m/s, and we ask how long time it would take to go 45 m? We start with the formula above and make the t the subject. This would give us
Substituting in our values will give us
As expected we get the previous value back.
But an equation of the type
can have two solutions, we have ****
So does the time of –3 s have any physical meaning? Let us extend the v/t-diagram to negative times. This gives us
The area to the left is as big as the area to the left, but positive. The time will now represent the time the object has to be thrown upwards to reach 45 m at the time 0 s, then to fall back down 45 m over the next 3 s.
An old example revisited
Let us look at the example from the first page about free fall. In that example we threw a ball straight up in the air with the velocity 20 m/s. An we got this graph:
I.e. an equation of a straight line with the “y“-intercept 20 and the slope or gradient –10.