Question

Can uh help in in this question step by step​

Answer 1

Answer . The acceleration of the truck is 10m/
`s^(2)`, and the distance covered is 40 m. Have attached the picture for solution.

Hope that helps.

Answer 2

• Initial velocity=u=72km/h

Convert to m/s

`\\ \sf \longmapsto 72* (5)/(18)=5(4)=20m/s`

• Final velocity=v=0m/s
• Time=2s=t

`\\ \sf \longmapsto Acceleration=(v-u)/(t)`

`\\ \sf \longmapsto Acceleration=(0-20)/(2)`

`\\ \sf \longmapsto Acceleration=(-20)/(2)`

`\\ \sf \longmapsto Acceleration=a=-10m/s^2`

• Distance be s

Using second equation of kinematics

`\\ \sf \longmapsto s=ut+(1)/(2)at^2`

`\\ \sf \longmapsto s=20(2)+(1)/(2)(-10)(2)^2`

`\\ \sf \longmapsto s=40+(-20)`

`\\ \sf \longmapsto s=40-20`

`\\ \sf \longmapsto s=20m`

Now

• Mass=m=5000kg

Using newtons second law

`\\ \sf \longmapsto Force=ma`

`\\ \sf \longmapsto Force=5000(-10)`

`\\ \sf \longmapsto Force=-50000N`

• Force is in opposite direction so its negative

`\\ \sf \longmapsto Force=50kN`