At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
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:
- Initial velocity would be [tex]u = 20\; {\rm m\cdot s^{-1}}[/tex].
- Final velocity would be [tex]0\; {\rm m\cdot s^{-1}}[/tex].
[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].
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.