according to recent typical test data, a Ford Focus travels 0.23 mi in 19 seconds, starting from rest. The same car, when braking from 59.0mph on dry pavement, stops in 141 ft. Assume constant acceleration in each part of its motion, but not necessarily the same acceleration when slowing down as when speeding up. Find the magnitude of this car's acceleration when braking.
first find the terminal velocity
distance=1/2 a t^2 solve for a
then, vf= at
Then solve for the braking acceleration
0= vabove*2 +2ad solve for a
To find the magnitude of the car's acceleration when braking, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 mph, as the car comes to a stop)
u = initial velocity (59.0 mph)
a = acceleration (which we need to find)
s = distance traveled (141 ft)
First, let's convert the initial velocity from mph to ft/s. There are 5280 ft in a mile and 3600 s in an hour.
59.0 mph * (5280 ft/mi) * (1/3600 h/s) = 86.6 ft/s (approx)
Now, we can plug in the values we have into the equation and solve for the acceleration (a):
0^2 = (86.6 ft/s)^2 + 2a * 141 ft
0 = 7500 ft^2/s^2 + 282a
Rearranging the equation:
-282a = 7500 ft^2/s^2
a = (7500 ft^2/s^2) / (-282)
a ≈ -26.6 ft/s^2 (approx)
Therefore, the magnitude of the car's acceleration when braking is approximately 26.6 ft/s^2.