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Derivative of [s=ut+.5at^2]

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Nov7-12, 04:41 AM
P: 5
Hey guys,
I'm currently in year 11 (Australia) and my physics class has recently started projectile motion. I noticed in class that the elementary differential calculus I've been learning in math could be applied to the questions we are working on. I'd been itching to try it out and today, after school, I did a question and (naturally) was able to solve it using what I had learned.

In the particular question I solved, it was necessary to find the derivative of the formula:
f(t) = ut + .5at2 , where f(t) = SHorizontal
Which I found to be:
f'(t) = u + at
Now, I realized that this was part of a formula that I had previously learnt:
v = u + at
Here the formula equates to velocity, but in the prior it equated to the slope of the tangent. Now I'm left wondering why this repetition exists and what (if any) the connection is.

Thanks for any help.
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Nov7-12, 09:13 AM
Sci Advisor
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P: 6,734
Velocity is defined as the rate of change of position with respect to time.

If you were to plot position as a function of time, velocity would be the slope of the position-time curve. Similarly, acceleration is the rate of change of velocity with respect to time. If you plotted velocity as a function of time, acceleration would be the slope of the velocity-time curve.
Nov7-12, 04:42 PM
P: 5
Ah, yes.
I realized before I fell asleep that the slope of the tangent at any point along that function with t on the x-axis is the velocity at that point in time (obviously).

Thanks for the help.

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