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a boss is moving with a velocity 72 km per hour on the seeing the rate signal of the traffic 27m ahead on the road,the driver applies brake and the bus stop at distance of 25m. Calculate the time taken by the car to come to rest​

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].