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