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A bus is going 30m.s-1 and stop in 5s.what is it's stopping distance for this speed​

Sagot :

  • Initial velocity=u=30m/s
  • Time=t=5s
  • Final velocity=v=0m/s.
  • Distance=s

[tex]\\ \tt\longmapsto Acceleration(a)=\dfrac{v-u}{t}[/tex]

[tex]\\ \tt\longmapsto a=\dfrac{0-30}{5}[/tex]

[tex]\\ \tt\longmapsto a=\dfrac{-30}{5}[/tex]

[tex]\\ \tt\longmapsto a=-6m/s^2[/tex]

Now using 2nd equation of kinematics

[tex]\\ \tt\longmapsto s=ut+\dfrac{1}{2}at^2[/tex]

[tex]\\ \tt\longmapsto s=30(5)+\dfrac{1}{2}(-6)(5)^2[/tex]

[tex]\\ \tt\longmapsto s=150+(-3)(25)[/tex]

[tex]\\ \tt\longmapsto s=150-75[/tex]

[tex]\\ \tt\longmapsto s=75m[/tex]

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