Kinematics moving particle problem

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The discussion revolves around a kinematics problem involving a particle's position described by the equation x=ct²-bt³. Participants are tasked with determining the units of constants c and b, concluding that c has units of m/s² and b has units of m/s³. The maximum positive x position occurs at t=1.0 seconds, and the user expresses uncertainty about using the quadratic formula for this cubic equation. The conversation highlights the need for a proper understanding of calculus to find relative maxima in such functions. Overall, the thread emphasizes the importance of unit analysis and calculus in solving kinematic problems.
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Homework Statement


The position of a particle moving along the x-axis depends on the time according to the equation

x=ct^{2}-bt^{3}

where x is in meters and t in seconds. What are the units of (a) constant c and (b) constant b? Let their numerical values be 3.0 and 2.0 respectively.(c) At what time does the particle reach its maximum positive x position? From t=0 to t=2.0, (d) what distance does the particle move and (e) what is its displacement? Find its velocity at times (f) 1.0 (g) 2.0 (h) 3.0 and (i)4.0. Find its acceleration at times (j) 1.0 (k) 2.0 (l) 3.0 and (m)4.0.

Homework Equations



x=ct^{2}-bt^{3}

The Attempt at a Solution



I can't even get started on part a. I am not sure how to find the constants. The only thing I could think of would be using the final unit (m), so in that case one would have to be meters and the other seconds?
Am I on the right track?
 
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You have x=ct2-bt3

'x' is in meters(m), so that the quantities ct2 and bt3 must also have units meters(m).

You know that 't' is in seconds (s), so 't2' has units s2, so what should the units of 'c' be such that (units of c)*(units of t2)= meters?
 
OK so for part A I have c=\frac{m}{s^{2}} and B as b=\frac{m}{s^{3}}. I believe the unit s should cancel out.
On part c I need to find a relative maximum. I can look at the graph and tell that t should equal 1. I was thinking to use the formula x=\frac{-b}{2a} then plug x in for t in the given equation, but I don't think I can use that because it is not a quadratic.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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