Respuesta :

  • Initial velocity=u=72km/h

Convert to m/s

[tex]\\ \sf \longmapsto 72\times \dfrac{5}{18}=5(4)=20m/s[/tex]

  • Final velocity=v=0m/s
  • Time=2s=t

[tex]\\ \sf \longmapsto Acceleration=\dfrac{v-u}{t}[/tex]

[tex]\\ \sf \longmapsto Acceleration=\dfrac{0-20}{2}[/tex]

[tex]\\ \sf \longmapsto Acceleration=\dfrac{-20}{2}[/tex]

[tex]\\ \sf \longmapsto Acceleration=a=-10m/s^2[/tex]

  • Distance be s

Using second equation of kinematics

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

[tex]\\ \sf \longmapsto s=20(2)+\dfrac{1}{2}(-10)(2)^2[/tex]

[tex]\\ \sf \longmapsto s=40+(-20)[/tex]

[tex]\\ \sf \longmapsto s=40-20[/tex]

[tex]\\ \sf \longmapsto s=20m[/tex]

Now

  • Mass=m=5000kg

Using newtons second law

[tex]\\ \sf \longmapsto Force=ma[/tex]

[tex]\\ \sf \longmapsto Force=5000(-10)[/tex]

[tex]\\ \sf \longmapsto Force=-50000N[/tex]

  • Force is in opposite direction so its negative

[tex]\\ \sf \longmapsto Force=50kN[/tex]

rubabzehra9127

Answer . The acceleration of the truck is 10m/[tex]s^{2}[/tex], and the distance covered is 40 m. Have attached the picture for solution.

Hope that helps.

Ver imagen rubabzehra9127
Ver imagen rubabzehra9127

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