A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 26 ft/s2. What is the distance covered before the car comes to a stop? (Round your answer to one decimal place.)
convert speeds to ft/sec
vf^2=vi^2+2ad solve for d
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To find the distance covered before the car comes to a stop, we can use the equation for motion with constant acceleration:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s since the car comes to a stop)
u = initial velocity (50 mi/h)
a = acceleration (-26 ft/s^2)
s = distance covered
First, let's convert the initial velocity from mi/h to ft/s:
u = 50 mi/h * 5280 ft/mi * 1/3600 h/s ≈ 73.33 ft/s
Next, we plug the given values into the equation:
0^2 = (73.33 ft/s)^2 + 2(-26 ft/s^2)s
Simplifying the equation:
0 = 5361.11 ft^2/s^2 - 52s
Rearranging the equation:
52s = 5361.11 ft^2/s^2
s = 5361.11 ft^2/s^2 / 52 ≈ 103 ft
Therefore, the car will cover approximately 103 feet before coming to a stop.