Since the truck was in motion before the brakes were applied, it will decelerate within the 4s before coming to a stop. Hence the acceleration is [tex]-a\:ms^{-2}[/tex].
When the truck comes to a stop, It will have a final velocity of [tex]v=0\:ms^{-1}[/tex].
Also the initial velocity is 22.0 m/s. This means, [tex]u=22.0\:ms^{-1}[/tex] and time, [tex]t=4s[/tex].
We can use the relation,
[tex]v=u+at[/tex]
to determine the acceleration of the truck.
Let us now plug in all the values to obtain,
[tex]0=22.0+(-a)(4)[/tex]
[tex]\Rightarrow -22.0=-4a[/tex]
[tex]\Rightarrow \frac{-22.0}{-4}=a[/tex]
[tex]\Rightarrow a=5.5 ms^{-2}[/tex]
Using the relation,
[tex]F=ma[/tex]
the magnitude of the braking force is
[tex]F=2000 \times 5.5 N=11000N[/tex]
