Finding the Constants in a Motion Equation

[fghtffyrdmns]

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)?

 

[Dick]

Integrate a(t) twice. You get two undetermined constants. Find them by using the s(0) and v(1) conditions.

 

[fghtffyrdmns]

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?

 

[PAR]

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

 

[fghtffyrdmns]

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

Nothing at all.

 

[SpicyPepper]

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.

 

[fghtffyrdmns]

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? :/

 

[PAR]

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?

 

[fghtffyrdmns]

Could you not do the inverse of the power rule?

 

[PAR]

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.

 

[fghtffyrdmns]

then s = 3qt^2 + 2bt + c?

 

[PAR]

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

 

[fghtffyrdmns]

I'm so lost right now.

My teacher shall get angry letters >:[

 

[PAR]

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.

 

[SpicyPepper]

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

 

[fghtffyrdmns]

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?

 

[PAR]

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

 

[fghtffyrdmns]

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?

 

[PAR]

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.

 

[SpicyPepper]

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)

 

[fghtffyrdmns]

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?

 

[fghtffyrdmns]

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

?

 

[SpicyPepper]

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.

 

[SpicyPepper]

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.

 

[fghtffyrdmns]

Why do I get 2 values for C, though?

 

[SpicyPepper]

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

 

[fghtffyrdmns]

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

t= 3/2 and -2

 

[SpicyPepper]

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

 

[fghtffyrdmns]

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.

 

[Dick]

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'?