How Does Differential Calculus Enhance Understanding of Projectile Motion?

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Differential calculus enhances the understanding of projectile motion by allowing students to analyze the relationships between position, velocity, and acceleration through derivatives. In projectile motion, the position function f(t) = ut + 0.5at² can be differentiated to find the velocity function f'(t) = u + at, illustrating that velocity is the slope of the position-time curve. This connection reveals that both the derivative and the velocity formula represent the same concept of rate of change. Additionally, acceleration, defined as the rate of change of velocity, can be understood similarly through its derivative. Overall, applying differential calculus to physics problems deepens comprehension of motion dynamics.
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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:
Code:
f(t) = ut + .5at[SUP]2[/SUP] , where f(t) = S[SUB]Horizontal[/SUB]
Which I found to be:
Code:
f'(t) = u + at
Now, I realized that this was part of a formula that I had previously learnt:
Code:
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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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.
 
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.
 
I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

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