– Student’s position:
\( x_s(t) = v_s t \)
\( v_s = 5.0 \, \text{m/s} \)

– Bus’s position:
\( x_b(t) = x_0 + \dfrac{1}{2} a_b t^2 \)
\( x_0 = 40.0 \, \text{m} \)
\( a_b = 0.170 \, \text{m/s}^2 \)

\( x_s(t) = x_b(t) \)
\( v_s t = x_0 + \dfrac{1}{2} a_b t^2 \)
Rearranged:
\( \dfrac{1}{2} a_b t^2 – v_s t + x_0 = 0 \)

\( \dfrac{1}{2} (0.170) t^2 – 5.0 t + 40.0 = 0 \)
Simplify coefficients:
\( 0.085 t^2 – 5.0 t + 40.0 = 0 \)

\( t = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
Where \( a = 0.085 \), \( b = -5.0 \), \( c = 40.0 \)
Discriminant \( D = b^2 – 4ac \):
\( D = (-5.0)^2 – 4(0.085)(40.0) = 25.0 – 13.6 = 11.4 \)
\( \sqrt{D} = \sqrt{11.4} \approx 3.376 \)
Solutions:
\( t = \dfrac{5.0 \pm 3.376}{0.17} \)

First solution:
\( t = \dfrac{5.0 + 3.376}{0.17} = \dfrac{8.376}{0.17} \approx 49.27 \, \text{s} \)

Second solution:
\( t = \dfrac{5.0 – 3.376}{0.17} = \dfrac{1.624}{0.17} \approx 9.553 \, \text{s} \)

Calculate the distance the student runs:
\( x_s = v_s t = 5.0 \times 9.553 \approx 47.77 \, \text{m} \)

\( v_b = a_b t = 0.170 \times 9.553 \approx 1.624 \, \text{m/s} \)

– Student’s path: Straight line starting from \( x = 0 \) with slope \( v_s = 5.0 \, \text{m/s} \).
– Bus’s path: Parabola starting from \( x = 40.0 \, \text{m} \) with increasing slope due to acceleration.

– This is when the bus overtakes the student again.
– Bus’s velocity at this time:
\( v_b = a_b t = 0.170 \times 49.27 \approx 8.376 \, \text{m/s} \)

\( 0.085 t^2 – 3.5 t + 40.0 = 0 \)
Discriminant \( D = (-3.5)^2 – 4(0.085)(40.0) = 12.25 – 13.6 = -1.35 \)
Since \( D < 0 \), no real solutions.

\( (-v_{s_{\text{min}}})^2 – 4(0.085)(40.0) = 0 \)
\( v_{s_{\text{min}}}^2 = 13.6 \)
\( v_{s_{\text{min}}} = \sqrt{13.6} \approx 3.692 \, \text{m/s} \)

\( t = \dfrac{v_{s_{\text{min}}}}{2a_b} = \dfrac{3.692}{2 \times 0.170} \approx 10.86 \, \text{s} \)
\( x_s = v_{s_{\text{min}}} t = 3.692 \times 10.86 \approx 40.0 \, \text{m} \)