Differentiation with physicsby JeffGlasgow Tags: acceleration, calculus, differentiation, motion, speed 

#1
May813, 02:03 AM

P: 5

1. The problem statement, all variables and given/known data
A moving point has a position function (P) and is given by 2D quantities where the (Y) axis is affected by gravity (9.8m/s/s) and (t) is in seconds. X(t)=4t cos θ and Y(t)= 4t sin (θ)5t^2 If θ = Pi /3 find the speed in directions after 10 seconds Find the angle where maximum X direction = maximum Y direction Find the time for maximum x direction = maximum Y direction Find the acceleration after 2 seconds Does the acceleration vary against time? Relevant equations/notes Equations of linear motion. 9.8m/s/s can be rounded to 10m/s/s for simplicity. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 



#2
May813, 02:43 AM

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P: 4,301

Hi Jeff, welcome to PF!
We have put the template that you haven't used in your post for a reason :) Maybe you can tell us what the "equations of linear motion" you mention are. I'm sure you already have some idea which one(s) to use? 



#3
May813, 03:06 AM

P: 5

After a bit of research I'm under the impression I need to use the formulae
V= u + at S = ut + 1/2 at^2 V^2 = u^2 + 2as Also my error in that the position function should be (s) instead of (p) 



#4
May813, 04:04 AM

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Differentiation with physics
Those formulas only hold for motion with constant acceleration a.
In your topic title you said "Differentiation with physics". What does differentiation have to do with position and velocity? 



#5
May813, 04:15 AM

P: 5

Our lecturer has just tried to make it relevant/more difficult due to the fact that we are doing an engineering course.




#6
May813, 05:00 AM

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Actually I just realised that the formulas you gave are applicable in this case, so I'm a bit confused now whether your lecturer wants you to use differentiation or not.
I'm tempted to advise you to try both ways, but let's go for the first one first, since you have found those formulas. What you should remember for 2D problems is that you need to consider the X and Y directions separately. So looking at Then what can you say about the formulas v_{x}, v_{y} for the velocity ? 



#7
May813, 05:43 AM

P: 5

I have no idea where to start with this one, let me try and find some notes on this....




#8
May813, 06:39 AM

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Look at the t's in the formula. The general expression for S(t) you gave is (something) * t + (something) * tē.
Can you write the expression for X(t) and Y(t) in the same way? Then you can compare the "somethings" and you should be able to read off what u and (1/2)a are. 



#9
May913, 01:31 AM

P: 5

For the y I get
S(t) = ut + 1/2 a t^2 Y(t) = 4t sin θ 5t^2 therefore I think that u =4 and a=10 but I'm not sure where the sin theta comes from And for the x I get S(t) = ut + 1/2 a t^2 X(t) = 4t cosθ Therefore I think that u =4 again but I have nothing for a?! And I'm unsure where cos is relevant? 


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