Finding the Constants in a Motion Equation

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The discussion focuses on solving the motion equation given the acceleration function a(t) = 24t + 6, with initial conditions s(0) = 0 and v(1) = 24. Participants emphasize the importance of integrating the acceleration function twice to derive the position function s(t) and the velocity function v(t). They clarify that the integration process yields undetermined constants, which can be solved using the provided initial conditions. The final forms of the functions are v(t) = 12t^2 + 6t + C and s(t) = 4t^3 + 3t^2 + Kt + D, where C and K are constants determined by the initial conditions.

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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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