Mechanics- general motion in a straight line

In summary, the conversation discusses the continuity of two functions, s(t) and v(t), and the values of different variables at specific intervals. There is also a question about finding the value of x, and a mention of a sudden drop in speed at t=4.
  • #1
Shah 72
MHB
274
0
20210604_200321.jpg

I don't know how to solve this
 
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  • #2
When are you going to start posting images that are reader friendly?

motion_prob10.jpg
 
  • #3
Both functions have to be continuous ...

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

$v(t)=\left\{\begin{matrix}
10t & t \in [0,4]\\
\frac{A}{2\sqrt{t}} +B& t \in (4,25]\\
C & t \in (25,50]
\end{matrix}\right.$
 
  • #4
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
 
  • #5
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!
 
  • #6
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
 
  • #7
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 …
 
  • #8
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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