Answer
$a. \quad 26$ seconds after Michelle starts.
$b.\quad $Pam catches and overtakes Michelle 5m before the finish line.
Work Step by Step
We start counting time after Michelle starts.
After t seconds,
Michelle will have been running for $t$ seconds for a distance of $x=7.5t$ meters.
Pam will have been running for ($t-1)$ seconds for a distance of $y=7.8(t-1)$ meters.
$a.$
To catch up, the distances covered are equal, $x=y.$
Substituting,
$7.5t=7.8(t-1)$
$ 7.5t=7.8t-7.8\qquad$ ... add $7.8-7.5t$
$7.8=0.3t\qquad $... divide with $0.3$
$t=\displaystyle \frac{7.8}{0.3}=26$ seconds after Michelle starts
$b.$
After $26$ seconds, Michelle will cover a distance of
$x=7.5(2.6)=195$ m
So, Pam catches and overtakes Michelle 5m before the finish line.
