Answer
$e=3.50$ m/s.
Work Step by Step
$\left[\begin{array}{lll}
\vec{r}=x\hat{\mathrm{i}}+y\hat{\mathrm{j}}+z\hat{\mathrm{k}} & (4- 1) & \text{Position vector} \\
\vec{v}=\frac{d\vec{r}}{dt} & (4- 10) & \text{instantaneous velocity} \\
\vec{v}=v_{x}\hat{\mathrm{i}}+v_{y}\hat{\mathrm{j}}+v_{z}\hat{\mathrm{k}}, & (4- 11) & \text{instantaneous velocity} \\
& & v_{x}=dx/dt, v_{y}=dy/dt,v_{z}=dz/dt.
\end{array}\right]$
Given $\quad \vec{r}=5.00t\hat{\mathrm{i}}+(et+ft^{2})\hat{\mathrm{j}},$
$v_{x}=dx/dt=5.00$ m/s
$v_{y}=dy/dt=(e+2ft)$ m/s
$\vec{v}=5.00\hat{\mathrm{i}}+(e+2ft)\hat{\mathrm{j}}$
The angle of $\vec{v} $ is
$\displaystyle \theta=\tan^{-1}(\frac{v_{y}}{v_{x}})$
$\displaystyle \theta=\tan^{-1}(\frac{e+2ft}{5.00})\qquad(*)$
(a)
Reading the graph at $t=0,$ we see $\theta=35.0^{\mathrm{o}}$.
Insert into the expression$\qquad(*)$
$35.0^{\mathrm{o}}=\displaystyle \tan^{-1}(\frac{e+2f(0)}{5.00})$
$\displaystyle \tan(35.0^{\mathrm{o}})=\frac{e}{5.00}$
$e=5.00\tan(35.0^{\mathrm{o}})=3.50$ m/s.
