Question

Can uh help in in this question step by step​

Answer 1

• Initial velocity=u=72km/h

Convert to m/s

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

• Final velocity=v=0m/s
• Time=2s=t

$\\ \sf \longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf \longmapsto Acceleration=\dfrac{0-20}{2}$

$\\ \sf \longmapsto Acceleration=\dfrac{-20}{2}$

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

• Distance be s

Using second equation of kinematics

$\\ \sf \longmapsto s=ut+\dfrac{1}{2}at^2$

$\\ \sf \longmapsto s=20(2)+\dfrac{1}{2}(-10)(2)^2$

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

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

$\\ \sf \longmapsto s=20m$

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

$\\ \sf \longmapsto Force=50kN$

Answer 2

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.