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.
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.
To determine the magnitude of acceleration, we can use the formula:
acceleration = (final velocity - initial velocity) / time
Given:
- Initial velocity (u) = 60 mi/h
- Final velocity (v) = 40 mi/h
- Time (t) = 3 s
First, let's convert the velocities from miles per hour to feet per second, as the time is in seconds and we need consistent units. We know that 1 mile is equivalent to 5280 feet, and 1 hour is equivalent to 3600 seconds.
Initial velocity (u) = 60 mi/h * (5280 ft/mi) / (3600 s/h) = 88 ft/s
Final velocity (v) = 40 mi/h * (5280 ft/mi) / (3600 s/h) = 58.67 ft/s
Now we can calculate the magnitude of acceleration:
acceleration = (v - u) / t
acceleration = (58.67 ft/s - 88 ft/s) / 3 s
acceleration = -29.33 ft/s / 3 s
acceleration = -9.78 ft/s²
Note: The negative sign indicates deceleration or slowing down.
Therefore, the magnitude of acceleration is approximately 9.78 ft/s².
Next, to find the distance traveled during the braking period, we can use the following formula:
distance = initial velocity * time + (1/2) * acceleration * time²
Considering that the initial velocity (u) is 88 ft/s, the time (t) is 3 s, and the acceleration (a) is -9.78 ft/s²:
distance = 88 ft/s * 3 s + (1/2) * (-9.78 ft/s²) * (3 s)²
distance = 264 ft + (-14.67 ft/s²) * 9 s²
distance = 264 ft + (-14.67 ft/s² * 81 s²)
distance = 264 ft - 1188.27 ft
distance ≈ -924.27 ft
The negative distance suggests that the car traveled in the opposite direction during the braking period. However, since distance is a scalar (magnitude without direction), we take the absolute value to ignore the direction. Therefore, the distance traveled during the braking period is approximately 924.27 feet.
To find the magnitude of the car's acceleration, we can use the formula:
acceleration = (final velocity - initial velocity) / time
Given:
Initial velocity (vi) = 60 mi/h
Final velocity (vf) = 40 mi/h
Time (t) = 3 s
First, we need to convert the velocities from mi/h to ft/s:
1 mile = 5280 feet
1 hour = 60 minutes = 60 seconds
initial velocity (vi) = 60 mi/h * 5280 ft/mi / 60 min/h / 60 s/min = 88 ft/s
final velocity (vf) = 40 mi/h * 5280 ft/mi / 60 min/h / 60 s/min = 58.67 ft/s
Now we can calculate the acceleration:
acceleration = (vf - vi) / t
acceleration = (58.67 ft/s - 88 ft/s) / 3 s
acceleration = -9.44 ft/s^2
The magnitude of the car's acceleration is 9.44 ft/s^2.
To find the distance traveled during the braking period, we can use the formula:
distance = initial velocity * time + (1/2) * acceleration * time^2
Substituting the given values:
distance = 88 ft/s * 3 s + (1/2) * (-9.44 ft/s^2) * (3 s)^2
distance = 264 ft - (1/2) * 9.44 ft/s^2 * 9 s^2
distance = 264 ft - 42.48 ft
distance = 221.52 ft
Therefore, the car travels a distance of 221.52 feet during the braking period.