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Physics differentiation: moving point has position function

JeffGlasgow

New member
Joined
May 9, 2013
Messages
2
A moving point has a position function (S) 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 1second

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....

Use Equations of linear motion. Eg.

V= u + at

S = ut + 1/2 at^2

V^2 = u^2 + 2as

9.8m/s/s can be rounded to 10m/s/s for simplicity
 
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stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,937
A moving point has a position function (S) 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 1second

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?
How far have you gotten in working through the steps of this exercise?

Please reply with your work so far, starting with your differentiations with respect to time "t". Thank you! ;)
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
4,716
A moving point has a position function (S) 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 1second

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....

Use Equations of linear motion. Eg.

V= u + at

S = ut + 1/2 at^2

V^2 = u^2 + 2as

9.8m/s/s can be rounded to 10m/s/s for simplicity
You seem to be saying that you have absolutely no idea what this is about! You could see that there is no "g" in this problem can't you? And you title this "Physics differentiation" so you must know that you can find the x and y coordinates of the velocity by differentiating the x and y coordinates of the position function. What do you get when you do that?
 
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