Question
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.
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.
Answers
GPT 3.5
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.
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.