When its brakes are applied, a certain automobile has constant deceleration of 22ft/s^2. If its initial velocity is 90 mi/h, how long will it take to come to a stop? How many feet will it travel during that time?
if acceleration is 22ft/s
then the formula for acceleration is
$\displaystyle g=-22$
now you need to find the velocity formula
where that is the anti derivative of acceleration
$\displaystyle v(t)=-22t+90$
where t is time in seconds and 90 is the inital velocity which in this case C
to show my work
$\displaystyle 90=-22(0)+C$
$\displaystyle C=90$
It will have stopped when velocity is 0
so $\displaystyle v(t)=-22t+90$ = 0 and solve for t
$\displaystyle 0=-22t+90$
$\displaystyle -90=-22t$
$\displaystyle 90/22=t$
t = 4.09
no you need to find the distance traveled for this time now you must take the anti derivative of the velocity formula which is
$\displaystyle s(t)=-11t^2+90t+C$
C is in the distance formula for something like this is 0 because C or s(t) sub 0 is usually initial height
so now plug in
t = 4.09
$\displaystyle s(4.09)=-11(4.09)^2+90(4.09)$
which equals 184.0909 feet
I maybe be totally wrong here
but it takes 4.09 seconds to stop and the car travels a distance of 184.09 feet
If your acceleration is in ft/s/s, than when you integrate, your velocity is in terms of ft/s. Meaning your constant of integration cannot be in mi/h.
Substitute 132ft/s for 90m/h. And do your time and distance calculations again. Everything else you did was right.