Answer:
[tex]2.5\; {\rm s}[/tex], assuming that acceleration would be constant when the brakes are applied.
Explanation:
Ensure that all quantities are in standard units. Initial velocity of the bus would be:
[tex]\displaystyle u = 72\; {\rm km\cdot h^{-1}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} = 20\; {\rm m\cdot s^{-1}}[/tex].
In a motion where acceleration is constant, average velocity during the entire motion would be equal to the average between initial velocity [tex]u[/tex] and final velocity [tex]v[/tex]:
[tex]\displaystyle (\text{average velocity}) = \frac{u + v}{2}[/tex].
In this question, assuming that acceleration of the bus is constant while the brake is applied, average velocity of the bus during the entire motion would be the average of the velocity before and after applying the brakes:
[tex]\begin{aligned} (\text{average velocity}) &= \frac{u + v}{2}\\ &= \frac{20\; {\rm m\cdot s^{-1}} + 0\; {\rm m\cdot s^{-1}}}{2} \\ &= 10\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
To find the duration of the motion, divide displacement by average velocity:
[tex]\begin{aligned}(\text{time required}) &= \frac{(\text{displacement})}{(\text{average velocity})} \\ &= \frac{25\; {\rm m}}{10\; {\rm m\cdot s^{-1}}} \\ &= 2.5\; {\rm s}\end{aligned}[/tex].
