- Thread starter
- #1

- Thread starter Shah 72
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- Thread starter
- #1

- Mar 1, 2012

- 935

- Mar 1, 2012

- 935

$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.$

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- #4

Iam so sorry. I thought it was OK. I will keep it in mind next time

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- #5

Thank you so much!!

$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.$

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- #6

I did q(a) as s is continues at t=4Thank you so much!!

80=2A+4B

Therefore A+2B= 40

Iam not getting the ans for q(d) Find the value of x

$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.$

- Mar 1, 2012

- 935

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 …

- Thread starter
- #8

Thank you!

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 …