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.