Please help,

Water decelerates in a diffuser following the relation $V=V_0(1-A\frac{x}{(L^2{)}^2})$ where $x$ is the distance from the start of the diffuser. At time $=0,$ the flow to the diffuser is turned off slowly and the inlet velocity $V_0$ decays with time as $V_0=V_i{e}^{-bt}.$ The velocity $V_i=29f\frac{t}{s},A=0\mathrm{.}7,L=2\mathrm{.}8ft,$ and $b=0\mathrm{.}1s^{-1}.$ The flow can be treated as one$-$dimensional.

Determine the local acceleration at time $=0$ and $5$ s at both the entrance and the exit.

At the entrance at $t=0,a_{\text{l}\text{o}\text{c}\text{a}\text{l}\text{,}0}=f\frac{t}{s^2}$

$2.$ Determine the convective acceleration at time $=0$ and $5$ s at both the entrance and the exit.

At the entrance at $t=0,a_{\text{c}\text{o}\text{n}\text{v}\text{,}0}=$

$f\frac{t}{s^2}$

$att=5s,a_{\text{c}\text{o}\text{n}\text{v}\text{.}0}=$

$f\frac{t}{s^2}$

$att=0,a_{\text{c}\text{o}\text{n}\text{v}\text{,}0}=$

$f\frac{t}{s^2}$

$att=0,a_{\text{c}\text{o}\text{n}\text{v}\text{.}2\mathrm{.}8ft}=$

$att=5s,a_{\text{c}\text{o}\text{n}\text{v}\text{.}2\mathrm{.}8ft}=$

$f\frac{t}{s^2}$

Determine the acceleration of the fluid at time $=0$ and $5$ s at both the entrance and the exit.

At the entrance at $t=0,a_0=f\frac{t}{s^2}$

at $t=5s,a_0=f\frac{t}{s^2}$

At the exit at $t=0,a_{2\mathrm{.}8ft}=f\frac{t}{s^2}$

at $t=5s,a_{2\mathrm{.}8ft}=f\frac{t}{s^2}$