Answer
$f=-0.125 m/s^{2}$
Work Step by Step
(b)
We have found so far::
$\displaystyle \theta=\tan^{-1}(\frac{e+2ft}{5.00})\qquad(*)$,
and $e=3.50$ m/s.
Reading the graph at $14.0$ s, we see $\theta=0^{\mathrm{o}}$.
Insert into the expression$\qquad(*)$, along with $e=3.50$ m/s we found in part (a).
$\displaystyle \theta=\tan^{-1}(\frac{e+2ft}{5.00})$
$\displaystyle \tan 0^{o}=\frac{3.50m/s+2f(14.0s)}{5.00}$
$0=3.50m/s+f(28.0s)$
$-3.50m/s=(28.0s)f$
$f=\displaystyle \frac{-3.50m/s}{28.0s}=-0.125 m/s^{2}$
