Question

A car slows down uniformly from a speed of 22.0 m/s to rest in 5.50 s. how far did it travel in that time?

Answer 1

Since the acceleration is 'uniform', the car's average speed during that time is 11 m/s.

Traveling at an average speed of 11 m/s for 5.5 sec, it covers

(11 m/s) x (5.5 sec) = 60.5 meters. (No calculus. Hardly any algebra. Mostly arithmetic.)

Answer 2

First get the acceleration.  Since it's uniform we use

$a= \frac{-22m/s}{5.5s}=-4 \frac{m}{s^2}$

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

$v=- \int\limits^{ } _ {}{4 \frac{m}{s^2} } \, dt =-4t \frac{m}{s^2}+22.0\frac{m}{s}$

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:

$s= \int\limits^{5.5} _0{-4t \frac{m}{s^2} +22.0\frac{m}{s} } \, dt =-2t^2\frac{m}{s^2} +22.0t\frac{m}{s} \\ \\ Evaluating: \\ \\ -2(5.5)^2+22(5.5)=60.5m}$

So it travels 60.5 meters