A car traveling at 72 mph starts braking and decelerates at a constant rate. the car takes 1/9 of a mile to stop. Find, to the nearest half-second, how long it took the car to cover the 1/9 of a mile.

I don't need the whole thing done. Just don't know how to start!

2 answers

LOL
there is a hard way and an easy way
the easy way is to know that during constant acceleration the average speed is the average of the beginning and end speeds

average speed = 72/2 = 36 mph

speed * time = distance
36 * time = 1/9
time = 1 hour/(9*36)
1 hour = 3600 seconds
so in seconds
3600/(9*36) = 11 seconds
now the hard way is
v = Vi + a t

0 = 72 + a t

a = -72/t

x = Vi t + .5 a t^2

1/9 = 72 t + .5 (-72/t) (t^2)

1/9 = 72 t - .5(72 t)
or like we said before
1/9 = (72/2) t