A car slows down from 18 m/s to rest in a distance of 91 m. What was its acceleration, assumed constant?

(Vf)^2 = Vo^2 + 2ad = 0,

(18)^2 + 2a*91 = 0,
324 + 182a = 0,
182a = -324,
a = -1.78m/s^2.

In coming to a stop, a car leaves skid marks 93 m long on the highway. Assuming a deceleration of 7.50 m/s2, estimate the speed of the car just before braking.

To find the acceleration of the car, we can use the first equation of motion:

v^2 = u^2 + 2as

where:
v = final velocity = 0 m/s (since the car comes to rest)
u = initial velocity = 18 m/s
s = distance traveled = 91 m

Rearranging the equation to solve for acceleration (a), we have:

a = (v^2 - u^2) / (2s)

Plugging in the values:

a = (0^2 - 18^2) / (2 * 91)

Simplifying:

a = (-324) / 182

a = -1.78 m/s^2

Therefore, the acceleration of the car is approximately -1.78 m/s^2, where the negative sign indicates that the car is decelerating.