It is given that
Speed of first van=40mph
Speed of second van=61mph
Distance=63 miles
We know that
![\text{Speed}=\frac{dis\tan ce}{\text{time}}](https://img.qammunity.org/2023/formulas/mathematics/college/4jb2q4vyicddwkb46ld40d9944r0apgmai.png)
![\text{time}=\frac{dis\tan ce}{\text{speed}}\ldots\text{.}\mathrm{}(1)](https://img.qammunity.org/2023/formulas/mathematics/college/kwglrqbf7w1etoxvj6pb6rqezgwjsi4nyn.png)
Substitute distance =63 miles and speed=40mph, we get the time taken by the professor
![\text{time}=(63)/(40)h](https://img.qammunity.org/2023/formulas/mathematics/college/onr9spygr3h7wlvixw887jcuy3nb6wvi0w.png)
![\text{time}=1.575h](https://img.qammunity.org/2023/formulas/mathematics/college/mzlagnwppju1ysa7wt9rljghlaoukis9ag.png)
Multiply by 60 to convert this time into minutes, we get
![\text{time}=1.575*60\text{minutes}](https://img.qammunity.org/2023/formulas/mathematics/college/7bslfboejzhk6jkyx42ywxvutxsaszvco6.png)
![\text{time}=94.5\text{minutes}](https://img.qammunity.org/2023/formulas/mathematics/college/zirscbx5r2ntqaxp2m4jk4sbt1mzursb6p.png)
94.5 minutes taken by the professor.
Substitute distance =63 miles and speed=61mph in equation (1), we get the time taken by the student
![\text{time}=(63)/(61)h](https://img.qammunity.org/2023/formulas/mathematics/college/nzcm5niwxa8si1x3bfi52wehliw4vkdhdq.png)
![\text{time}=1.0328h](https://img.qammunity.org/2023/formulas/mathematics/college/5gcl2hwp4sml2y7mbo3av4ua13ytrbgt1u.png)
Multiply by 60 to convert this time into minutes, we get
![\text{time}=1.0328*60\text{ minutes}](https://img.qammunity.org/2023/formulas/mathematics/college/wn7ph0rwv9q1q6ahlq1qggeqp0fpuwxld6.png)
![\text{time}=61.9\text{ minutes}](https://img.qammunity.org/2023/formulas/mathematics/college/72emt5hyg3dwk1f07wcolhe9jaypldc0zj.png)
61.9 minutes taken by the student.