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= (-22m/s)/(5.5s)=-4 (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 (m)/(s^2) } \, dt =-4t (m)/(s^2)+22.0(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 (m)/(s^2) +22.0(m)/(s) } \, dt =-2t^2(m)/(s^2) +22.0t(m)/(s) \\ \\ Evaluating: \\ \\ -2(5.5)^2+22(5.5)=60.5m}`

So it travels 60.5 meters