She overtakes the bus??

Let's place the student and the bus in the $xy-$ plane such that the student is at the point $(-40,0)$ , and the bus stop is at the origin. Both entities move along the x-axis in the positive direction.

After $t$ seconds, the student is at $(-40+5t,0)$ and the bus at $(\frac{1}{2} at^2, 0)$ , or $(\frac{0.17}{2} t^2, 0).$ The student catches the bus when:

$-40 + 5t = \frac{0.17}{2} t^2 \Rightarrow 0 = 0.17t^2 - 10t + 80 \Rightarrow t = \frac{10 \pm \sqrt{10^2 - 4(0.17)(80)}}{0.34} \Rightarrow t = 9.55, 49.273$ seconds.

Since $9.55 < 49.273$ , this is the least time it takes the student to catch bus (i.e. Choice C).