Finding the Constants in a Motion Equation

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Homework Help Overview

The problem involves finding the position function s(t) given the acceleration function a(t) = 24t + 6, along with initial conditions for position and velocity. The subject area is calculus, specifically focusing on motion equations and derivatives.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss integrating the acceleration function to find velocity and position, while some express uncertainty about integration and its concepts. There are attempts to relate the derivatives of the functions to the given conditions, and questions arise about the constants involved in the equations.

Discussion Status

The discussion is ongoing with various approaches being explored. Some participants have offered guidance on using integration and the relationships between position, velocity, and acceleration. There is a recognition of the need to clarify the constants involved in the equations, and some participants are attempting to derive expressions for velocity and position based on the initial conditions.

Contextual Notes

Participants note that they have not yet learned integration, which complicates their ability to solve the problem using standard methods. There is also mention of the teacher assigning questions that have not been covered in class, contributing to the confusion.

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Homework Statement



If s(0) = 0, v(1) = 24 and a(t) = 24t+6 find s(t)

Homework Equations





The Attempt at a Solution



I know a(t) is s''(t) and v(t) is s'(t). however, How can I find s(t)?
 
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Integrate a(t) twice. You get two undetermined constants. Find them by using the s(0) and v(1) conditions.
 


We have not learned integration yet :(. Infact, we have not even done anything from finding a function from it's derivatives.

Is there any other way to do it?
 


Have you been given any equation relating s,v, and a?
 


PAR said:
Have you been given any equation relating s,v, and a?

Nothing at all.
 


Find the anti-derivative of a(t), which is v(t).

Plug in t=1 for v(t) to get the constant.

The find the anti-derivative of v(t), which is s(t).

Plug in t=0 for s(t) to get the constant, and now you have your answer.
 


SpicyPepper said:
Find the anti-derivative of a(t), which is v(t).

Plug in t=1 for v(t) to get the constant.

The find the anti-derivative of v(t), which is s(t).

Plug in t=0 for s(t) to get the constant, and now you have your answer.

What is this anti-derivative? What's integration? :/
 


Well either you haven't been paying attention in class or there is something seriously wrong with how they are teaching you.

Since you haven't learned integration, the only possible way I can see you finding s(t) by derivatives only is knowing that since a(t) = 24t + 6, all of the integrals of a(t) must obey the power rule. Knowing this you know that

v(t) = f(t^2) and s = g(t^3)

So g(t^3) is a cubic. and can be written in the form:

q*t^3 + b*t^2 + c*t + d

can you solve from there?
 


Could you not do the inverse of the power rule?
 
  • #10


The inverse of the power rule is integration. Since you know that

s = q*t^3 + b*t^2 + c*t + d

all you need to do is solve for q, b,c and d using derivatives.
 
  • #11


then s = 3qt^2 + 2bt + c?
 
  • #12


no, s = q*t^3 + b*t^2 + c*t + d
 
  • #13


I'm so lost right now.

My teacher shall get angry letters >:[
 
  • #14


Recap: You know that s(t) is a cubic because when you take two derivatives of a cubic you get At + B which is the same form as a(t) = 24t + 6.

So using the initial conditions eg. s(0) = 0, the a(t) formula, s(t) = q*t^3 + b*t^2 + c*t + d, and knowing that s' = v and s'' = a, solve for q,b,c, and d.
 
  • #15


Someone's been skipping class. Anti-derivative is basically doing the reverse of a derivative.

For example the derivative of 3x^2 + 5x + 4
= 6x + 5

The anti-derivative of 6x + 5
= 6x^2 * (1/2) + 5x * (1/1) + C
= 3x^2 + 5x + C

If the f(x) = 3x^2 + 5x + C
and we're given f(0)=4
then we can figure out the constant C
and we get f(x) = 3x^2 + 5x + 4

Hey guys, what's math? :p
 
  • #16


I think I got it! My book doesn't even have this stuff in it. No, I was not skipping class. This teacher just gives us questions we've not even seen before. He does it all the time.

v(t) = 12t^2+6t
s(t) = 4t^3+3t^2 + t?

no?
not even close?
 
  • #17


almost, v(t) is incorrect, redo the derivative of s(t) to find the correct v(t)
 
  • #18


PAR said:
almost, v(t) is incorrect, redo the derivative of s(t) to find the correct v(t)

am i missing the constant in the v(t)? a K right?
 
  • #19


Your s(t) is correct, now take its derivative to find v(t).

yes, you are missing a constant in v(t)

EDIT: Sorry I said that your s(t) is correct, sorry it isn't, the "t" term is wrong, you need a constant coefficient, so you need a Kt not a t. But using the initial conditions given to you, you can find what K is.
 
  • #20


fghtffyrdmns said:
v(t) = 12t^2+6t

You need the constant, so it's
v(t) = 12t^2+6t+C
to find C, you plug in the fact that you know v(1)=24

After you find C, then you repeat the process to find s(t)
 
  • #21


See, this is what I don't understand.

would I just write 24 = 12t^2+6t +c?

the solve for c? where does the K come from?
 
  • #22


v(t) = 12t^2+6t + C
s(t) = 4t^3+3t^2 + Kt +C

?
 
  • #23


fghtffyrdmns said:
See, this is what I don't understand.

would I just write 24 = 12t^2+6t +c?

the solve for c? where does the K come from?
If I'm understanding your use of variables correctly, K is C.

K is one of the coefficients in s(t), which is known by figuring out C in v(t).

v(t) = ... + C
s(t) = ... + Kt + ...

You wrote v(t) correctly, now just plug in t=1, and you'll get C.
 
  • #24


fghtffyrdmns said:
v(t) = 12t^2+6t + C
s(t) = 4t^3+3t^2 + Kt +C

?
yes, but C is not the same value or variable in both of those equations. C is just commonly used to represent an unknown constant.

The C in your first equation will be the K in your 2nd equation.
 
  • #25


Why do I get 2 values for C, though?
 
  • #26


You don't. You just reused the same variable.

I rewrote it:
v(t) = 12t^2+6t + C
s(t) = 4t^3+3t^2 + Ct + D

v(1)=24
s(0)=0
 
  • #27


Wait, I think I solved for C improperly. 24 = 12t^2 +6t + C
C = -12t^2-6t+24

t= 3/2 and -2
 
  • #28


v(1)=24, which means t = 1. Plug in 1 wherever you have a t.
 
  • #29


ahhh yes. I was thinking of something else sorry.

I got c at 6. which makes sense since d should be zero as s(0) = 0.
 
  • #30


fghtffyrdmns said:
I think I got it! My book doesn't even have this stuff in it. No, I was not skipping class. This teacher just gives us questions we've not even seen before. He does it all the time.

v(t) = 12t^2+6t
s(t) = 4t^3+3t^2 + t?

no?
not even close?

Close. The 't' is s(t) would give you a '1' in v(t). I don't see any. How about replacing it with a constant 'C'?
 

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