Solve V(t) from V(s): Steps & Answers

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Homework Help Overview

The problem involves finding the velocity function V(t) from a given velocity function V(s) with specific initial conditions. The context is rooted in differential equations and the relationship between position and time.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to derive V(t) by manipulating the differential relationships between v, s, and t. Some participants question the placement of constants in the equations, leading to discussions about algebraic manipulation and the implications of rearranging terms.

Discussion Status

The discussion has seen various interpretations of the algebra involved, with participants providing feedback on the original poster's approach. There is acknowledgment of potential algebra mistakes, and while some guidance has been offered, no explicit consensus on the correct form of V(t) has been reached.

Contextual Notes

Participants are navigating through algebraic expressions and constants, with some noting the importance of proper placement in equations. The original poster's initial conditions and the online system's feedback are also influencing the discussion.

Lancelot59
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I'm given this:
[tex] v=-4s^{2}[/tex]
s=2
t=0

and I need to find v(t)

I used this differential and got t like so:
[tex] dt=\frac{v}{ds}[/tex]

[tex] dt=\frac{ds}{v}[/tex]
[tex] dt=\frac{ds}{-4s^{2}}[/tex]
[tex] t=-\frac{1}{4s}+c[/tex]

I plugged in 0 for t and 2 for s. I got -1/8 for c. Then isolating s I got
[tex] s=-\frac{1}{4t}+2[/tex]
I stuck that into v=-4s^2 and got this:
[tex] v=-\frac{1}{4t^2}-\frac{4}{t}-16[/tex]
However according to the online system I'm wrong.
 
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Hi Lancelot59! :smile:

Your brackets are wrong :redface:

it's s = 1/(4t + constant) :wink:
 
How does that work? I thought constants were always added to the end.
 
Yes, which is where you correctly put it in your line t = 1/4s + C.

But it doesn't stay at the end when you up-end it …

1/4s = t - C,

4s = 1/(t - C) :wink:
 
I see, so now I get c=1/8.
I still end up with the same expression for S.

Nevermind, I made an algebra mistake. I have the answer now.

Thanks.
 
Last edited:

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