To find the magnitude of the car's acceleration, we can use the formula:
a = (vf - vi) / t
where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time.
Given:
Initial velocity, vi = 60 mi/h
Final velocity, vf = 40 mi/h
Time, t = 3 s
First, we convert the velocities from mi/h to ft/s:
vi = 60 mi/h * 5280 ft/mi * 1/3600 h/s = 88 ft/s
vf = 40 mi/h * 5280 ft/mi * 1/3600 h/s = 58.7 ft/s
Now, we can calculate the magnitude of the acceleration:
a = (58.7 ft/s - 88 ft/s) / 3 s
a = -29.3 ft/s / 3 s
a ≈ -9.77 ft/s²
The magnitude of the acceleration is approximately 9.77 ft/s².
To find the distance traveled during the braking period, we can use the formula:
d = vit + (1/2)at²
where d is the distance, vi is the initial velocity, t is the time, a is the acceleration, and t is the time.
Given:
Initial velocity, vi = 88 ft/s
Time, t = 3 s
Acceleration, a = -9.77 ft/s²
Plugging in these values, we get:
d = 88 ft/s * 3 s + (1/2)(-9.77 ft/s²)(3 s)²
d = 264 ft + (1/2)(-9.77 ft/s²)(9 s²)
d = 264 ft + (-44.07 ft)
d = 219.93 ft
The car travels approximately 219.93 ft during the braking period.
The driver of a car traveling on the highway suddently slames on the brakes of a slowdown in traffic ahead. If the car's speed decreases at a constant rate from 60mi/h to 40mi/h in 3s.
What is the magnitude of its acceleration, assuming it constinues to move in a straight line?
What distance does the car travel during the braking period?
Express answer in feet.
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