Exploring the Relationship Between Projectile Motion and Differential Calculus

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In summary, the conversation discusses the application of elementary differential calculus to projectile motion questions in a physics class. The speaker was able to solve a question using the derivative of a formula they had previously learned, and they noted the connection between velocity and the slope of the tangent. The conversation ends with the speaker thanking for any help.
  • #1
Fourthkind
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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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  • #2
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.
 
  • #3
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.
 

FAQ: Exploring the Relationship Between Projectile Motion and Differential Calculus

What is the formula for finding the derivative of [s=ut+.5at^2]?

The formula for finding the derivative of [s=ut+.5at^2] is d/dt(s)=u+at, where d/dt represents the rate of change over time.

Why do we need to find the derivative of [s=ut+.5at^2]?

Finding the derivative of [s=ut+.5at^2] allows us to calculate the instantaneous rate of change of an object's position over time. This is useful in many fields such as physics, engineering, and economics.

What does each variable in [s=ut+.5at^2] represent?

In this equation, s represents the position of the object, u represents the initial velocity, a represents the acceleration, and t represents time.

How do we find the derivative of [s=ut+.5at^2]?

To find the derivative of [s=ut+.5at^2], we use the power rule, which states that for any term with a variable raised to a power, we can bring the power down and multiply it by the coefficient, then subtract 1 from the original power. In this case, we would bring down the power of t in ut and at^2, resulting in u+at.

Can the derivative of [s=ut+.5at^2] be negative?

Yes, the derivative of [s=ut+.5at^2] can be negative. This would indicate that the object is decelerating, or slowing down, over time.

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