A 745i BMW car can brake to a stop in a distance of 122 ft from a speed of 60.0 mi/h. To brake to a stop from a speed of 80.0 mi/h requires a stopping distance of 221 ft. What is the average braking acceleration for the following intervals? Express the answers in mi/h/s and in m/s^2
60 mi/h to rest, 80 mi/h to rest,80 mi/h to 60 mi/h
Vf^2 = Vo^2 + 2ad
0 = Vo^2 + 2ad
Vo^2 = -2ad
At 60 mph:
V^2 = -2ad => a = -V^2/2d
a = (60.0 mi/h)^2 / (2)(0.0231mi)
a = -77901.64 mi/h^2
1 mi/h^2 = 3600 mi/(h*s) (why?)
1 mi/(h*s) = 1609m/(3600s)(1s) = 0.447 m/s^2
a = -21.64 mi/h/s = -9.67 m/s^2
You calculate the acceleration when Vo = 80 mph in a similar fashion.
To calculate the average braking acceleration, we will use the following formula:
Average acceleration = (Final velocity^2 - Initial velocity^2) / (2 * distance)
First, let's calculate the average braking acceleration for braking from 60 mi/h to rest:
Initial velocity = 60 mi/h
Final velocity = 0 mi/h
Distance = 122 ft
Converting the distance from feet to miles:
Distance = 122 ft * (1 mile / 5280 ft) = 0.0231 miles
Using the formula:
Average acceleration = (0 - 60^2) / (2 * 0.0231)
Average acceleration = (-3600) / (0.0462)
Average acceleration ≈ -77967.53 mi/h/s
To convert mi/h/s to m/s^2, we need to multiply by the conversion factor (1.60934 km/h * 1000 m/1 km / 3600 s/h):
Average acceleration ≈ -77967.53 mi/h/s * (1.60934 km/h * 1000 m/1 km / 3600 s/h)
Average acceleration ≈ -34526.87 m/s^2
Next, let's calculate the average braking acceleration for braking from 80 mi/h to rest:
Initial velocity = 80 mi/h
Final velocity = 0 mi/h
Distance = 221 ft
Converting the distance from feet to miles:
Distance = 221 ft * (1 mile / 5280 ft) = 0.0419 miles
Using the formula:
Average acceleration = (0 - 80^2) / (2 * 0.0419)
Average acceleration = (-6400) / (0.0838)
Average acceleration ≈ -76270.02 mi/h/s
Converting mi/h/s to m/s^2:
Average acceleration ≈ -76270.02 mi/h/s * (1.60934 km/h * 1000 m/1 km / 3600 s/h)
Average acceleration ≈ -33893.20 m/s^2
Lastly, let's calculate the average braking acceleration for braking from 80 mi/h to 60 mi/h:
Initial velocity = 80 mi/h
Final velocity = 60 mi/h
Distance = 221 ft - 122 ft = 99 ft
Converting the distance from feet to miles:
Distance = 99 ft * (1 mile / 5280 ft) = 0.01875 miles
Using the formula:
Average acceleration = (60^2 - 80^2) / (2 * 0.01875)
Average acceleration = (3600 - 6400) / (0.0375)
Average acceleration ≈ -106666.67 mi/h/s
Converting mi/h/s to m/s^2:
Average acceleration ≈ -106666.67 mi/h/s * (1.60934 km/h * 1000 m/1 km / 3600 s/h)
Average acceleration ≈ -47411.74 m/s^2
Therefore, the average braking accelerations for the given intervals are:
60 mi/h to rest: -77967.53 mi/h/s or -34526.87 m/s^2
80 mi/h to rest: -76270.02 mi/h/s or -33893.20 m/s^2
80 mi/h to 60 mi/h: -106666.67 mi/h/s or -47411.74 m/s^2
To find the average braking acceleration for each interval, we can use the following equation:
\[ a = \frac{v_f - v_i}{t} \]
Where:
- \( a \) is the acceleration,
- \( v_f \) is the final velocity,
- \( v_i \) is the initial velocity, and
- \( t \) is the time interval.
First, let's convert the speeds from miles per hour (mi/h) to meters per second (m/s) since the formula requires consistent units.
1 mile = 1609.34 meters
1 hour = 3600 seconds
So, to convert from mi/h to m/s, we can multiply the value by 1609.34/3600.
Using this conversion factor, we get:
60 mi/h = (60 * 1609.34) / 3600 m/s = 26.82 m/s
80 mi/h = (80 * 1609.34) / 3600 m/s = 35.76 m/s
Now, let's calculate the average acceleration for each interval.
1. 60 mi/h to rest:
The initial velocity, \( v_i \), is 26.82 m/s, and the final velocity, \( v_f \), will be 0 m/s since the car comes to rest. The time interval, \( t \), is not given directly, but we can calculate it.
Using the formula for acceleration, \( a \):
\( a = \frac{v_f - v_i}{t} \)
Rearranging the formula:
\( t = \frac{v_f - v_i}{a} \)
Substituting the values:
\( t = \frac{0 - 26.82}{a} \)
The stopping distance is given as 122 ft, which we need to convert to meters since the unit of acceleration is m/s^2.
1 foot = 0.3048 meters
So, the stopping distance is 122 * 0.3048 = 37.1864 meters.
Now we can calculate the acceleration:
\( a = \frac{0 - 26.82}{\frac{37.1864}{2}} \)
Simplifying:
\( a = \frac{-2 * 26.82}{37.1864} \)
Finally, converting the answer to m/s^2 by dividing by 1609.34:
\( a = \frac{-53.64}{37.1864} \) m/s^2
2. 80 mi/h to rest:
Using similar steps as above:
The initial velocity, \( v_i \), is 35.76 m/s, the final velocity, \( v_f \), is 0 m/s, and the stopping distance is 221 ft.
In this case:
\( t = \frac{0 - 35.76}{a} \)
\( a = \frac{-2 * 35.76}{\frac{221*0.3048}{2}} \)
\( a = \frac{-71.52}{67.5592} \) m/s^2
3. 80 mi/h to 60 mi/h:
Here, we know both the initial velocity and final velocity, but we don't know the time interval. However, we can calculate it using the stopping distance.
The stopping distance is 221 ft, which we convert to meters: 221 * 0.3048 = 67.3592 meters.
The change in velocity, \( \Delta v \), is \( v_f - v_i \):
\( \Delta v = 60 - 35.76 \) m/s
Using the formula for acceleration, \( a = \frac{\Delta v}{t} \), we can rearrange it to find \( t \):
\( t = \frac{\Delta v}{a} \)
Substituting the values:
\( t = \frac{60 - 35.76}{\frac{67.5592}{2}} \)
Simplifying:
\( t = \frac{24.24}{33.7796} \) s
Finally, we can divide the change in velocity by the calculated time interval to obtain the acceleration:
\( a = \frac{24.24}{33.7796} \) m/s^2
Therefore, the average braking accelerations for the given intervals are:
60 mi/h to rest: approximately -0.8037 m/s^2 or -1.4423 mi/h/s
80 mi/h to rest: approximately -1.0642 m/s^2 or -1.9056 mi/h/s
80 mi/h to 60 mi/h: approximately 0.7185 m/s^2 or 1.2889 mi/h/s