Mechanics- general motion in a straight line

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Discussion Overview

The discussion revolves around the continuity of piecewise functions representing position and velocity in the context of general motion in a straight line. Participants are exploring the implications of continuity at specific points and attempting to solve related questions.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that both functions must be continuous, particularly at the transition points between the defined intervals.
  • One participant expresses a concern about a discontinuity in the speed function at t = 4, suggesting that this affects the continuity of the velocity function.
  • Another participant presents a derived equation, A + 2B = 40, based on the continuity condition at t = 4.
  • There is a question about whether the adjustments made to the velocity function lead to a new form of the position function for the interval t ∈ (25,50].
  • Several participants express uncertainty about finding specific values or solutions related to the posed questions.

Areas of Agreement / Disagreement

Participants generally agree on the necessity of continuity for the functions involved, but there is disagreement regarding the implications of the discontinuity at t = 4 and how it affects the overall problem. The discussion remains unresolved with multiple competing views on the continuity and its consequences.

Contextual Notes

Limitations include potential missing assumptions regarding the definitions of continuity and the behavior of the functions at the transition points. The discussion does not resolve the mathematical steps needed to fully address the posed questions.

Shah 72
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20210604_200321.jpg

I don't know how to solve this
 
Mathematics news on Phys.org
When are you going to start posting images that are reader friendly?

motion_prob10.jpg
 
Both functions have to be continuous ...

$s(t)=\left\{\begin{matrix}
5t^2 &t\in [0,4] \\
A\sqrt{t}+Bt & t\in (4,25]\\
Ct+30 & t \in (25,50]
\end{matrix}\right.$

$v(t)=\left\{\begin{matrix}
10t & t \in [0,4]\\
\frac{A}{2\sqrt{t}} +B& t \in (4,25]\\
C & t \in (25,50]
\end{matrix}\right.$
 
skeeter said:
When are you going to start posting images that are reader friendly?

View attachment 11175
Iam so sorry. I thought it was OK. I will keep it in mind next time
 
skeeter said:
Both functions have to be continuous ...

$s(t)=\left\{\begin{matrix}
5t^2 &t\in [0,4] \\
A\sqrt{t}+Bt & t\in (4,25]\\
Ct+30 & t \in (25,50]
\end{matrix}\right.$

$v(t)=\left\{\begin{matrix}
10t & t \in [0,4]\\
\frac{A}{2\sqrt{t}} +B& t \in (4,25]\\
C & t \in (25,50]
\end{matrix}\right.$
Thank you so much!
 
Shah 72 said:
Thank you so much!
I did q(a) as s is continues at t=4
80=2A+4B
Therefore A+2B= 40
skeeter said:
Both functions have to be continuous ...

$s(t)=\left\{\begin{matrix}
5t^2 &t\in [0,4] \\
A\sqrt{t}+Bt & t\in (4,25]\\
Ct+30 & t \in (25,50]
\end{matrix}\right.$

$v(t)=\left\{\begin{matrix}
10t & t \in [0,4]\\
\frac{A}{2\sqrt{t}} +B& t \in (4,25]\\
C & t \in (25,50]
\end{matrix}\right.$
Iam not getting the ans for q(d) Find the value of x
 
I wasn’t able to find it either …

The sudden “drop” in speed at t = 4 makes the speed function discontinuous there.

$s’(t) = \dfrac{A}{2\sqrt{t}} + B$ becomes $s’(t) = \dfrac{A}{2\sqrt{t}} + B - x$

now, does that make $s(t) = (C-x)t + 30 \text{ and } s’(t) = C-x$ for $t \in (25,50]$ ?

Maybe I’m missing something …
 
skeeter said:
I wasn’t able to find it either …

The sudden “drop” in speed at t = 4 makes the speed function discontinuous there.

$s’(t) = \dfrac{A}{2\sqrt{t}} + B$ becomes $s’(t) = \dfrac{A}{2\sqrt{t}} + B - x$

now, does that make $s(t) = (C-x)t + 30 \text{ and } s’(t) = C-x$ for $t \in (25,50]$ ?

Maybe I’m missing something …
Thank you!
 

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