Shah 72
MHB
- 274
- 0
I don't know how to solve this
Iam so sorry. I thought it was OK. I will keep it in mind next timeskeeter said:
Thank you so much!skeeter said:Both functions have to be continuous ...
$s(t)=\left\{\begin{matrix}
5t^2 &t\in [0,4] \\
A\sqrt{t}+Bt & t\in (4,25]\\
Ct+30 & t \in (25,50]
\end{matrix}\right.$
$v(t)=\left\{\begin{matrix}
10t & t \in [0,4]\\
\frac{A}{2\sqrt{t}} +B& t \in (4,25]\\
C & t \in (25,50]
\end{matrix}\right.$
I did q(a) as s is continues at t=4Shah 72 said:Thank you so much!
Iam not getting the ans for q(d) Find the value of xskeeter 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!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 …