Mechanics- general motion in a straight line

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SUMMARY

The discussion focuses on the continuity of piecewise functions representing position and velocity in a straight line motion problem. The position function is defined as $s(t)=\left\{\begin{matrix} 5t^2 &t\in [0,4] \\ A\sqrt{t}+Bt & t\in (4,25]\\ Ct+30 & t \in (25,50] \end{matrix}\right.$ and the velocity function as $v(t)=\left\{\begin{matrix} 10t & t \in [0,4]\\ \frac{A}{2\sqrt{t}} +B& t \in (4,25]\\ C & t \in (25,50] \end{matrix}\right.$. Key insights include the need for continuity at transition points, specifically at $t=4$, where the velocity function experiences a discontinuity. The equation $80=2A+4B$ is derived to ensure continuity at this point.

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  • Familiarity with continuity concepts in mathematics
  • Ability to solve algebraic equations
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  • Study the concept of continuity in piecewise functions
  • Learn about derivatives and their applications in motion problems
  • Explore algebraic techniques for solving systems of equations
  • Investigate the implications of discontinuities in physical motion
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Shah 72
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20210604_200321.jpg

I don't know how to solve this
 
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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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