First get the acceleration. Since it's uniform we use
[tex]a= \frac{-22m/s}{5.5s}=-4 \frac{m}{s^2} [/tex]
Note the acceleration is negative since we start at a positive speed and end at zero.
Now the distance is the acceleration integrated twice. The first integral gives the velocity at any time, t
[tex]v=- \int\limits^{ } _ {}{4 \frac{m}{s^2} } \, dt =-4t \frac{m}{s^2}+22.0\frac{m}{s}[/tex]
Notice if you put 5.5s for t in the expression we get 0 m/s as we should. Now to get the distance it traveled over this time we integrate this velocity expression:
[tex]s= \int\limits^{5.5} _0{-4t \frac{m}{s^2} +22.0\frac{m}{s} } \, dt =-2t^2\frac{m}{s^2} +22.0t\frac{m}{s} \\ \\ Evaluating: \\ \\ -2(5.5)^2+22(5.5)=60.5m} [/tex]
So it travels 60.5 meters
