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?
then the formula for acceleration is
now you need to find the velocity formula
where that is the anti derivative of acceleration
where t is time in seconds and 90 is the inital velocity which in this case C
to show my work
It will have stopped when velocity is 0
so = 0 and solve for 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
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
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.