A car is traveling at 60 mi/h down a highway. a)What magnitude of acceleration does it need to have to come to a complete stop in a distance of 200 ft? b)What acceleration does it need to stop in 200 ft if it is traveling at 110 mi/h ?
60 mph = 88 ft/s
a) the average speed while stopping is 30 mph or 44 ft/s
... the stopping time is
... 200 ft / 44 ft/s
divide the initial speed by the stopping time to find the acceleration
b) same process as a) but with a different initial speed
To find the acceleration required for the car to come to a complete stop in a certain distance, we can use the equations of motion.
a) To find the magnitude of acceleration required for the car traveling at 60 mi/h to stop in 200 ft:
Step 1: Convert the speed from miles per hour to feet per second.
Conversion factor: 1 mile = 5280 feet, 1 hour = 3600 seconds
Speed in ft/s = (60 mi/h) * (5280 ft/mi) / (3600 s/h) = 88 ft/s
Step 2: Use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 ft/s in this case since the car comes to a stop)
u = initial velocity (88 ft/s)
a = acceleration (to be determined)
s = distance traveled (200 ft)
Rearrange the equation to solve for acceleration:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (0^2 - 88^2) / (2 * 200)
a = -7744 / 400
a = -19.36 ft/s^2
So, the magnitude of acceleration required for the car to come to a stop in 200 ft is 19.36 ft/s^2.
b) To find the acceleration required for the car traveling at 110 mi/h to stop in 200 ft:
Step 1: Convert the speed from miles per hour to feet per second.
Speed in ft/s = (110 mi/h) * (5280 ft/mi) / (3600 s/h) = 161.33 ft/s
Step 2: Use the same equation of motion:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (0^2 - 161.33^2) / (2 * 200)
a = -26,009.8464 / 400
a = -65.02 ft/s^2
So, the acceleration required for the car traveling at 110 mi/h to stop in 200 ft is approximately 65.02 ft/s^2.
To find the magnitude of acceleration required for the car to come to a complete stop in a given distance, we can use the equations of motion. The relevant equation in this case is:
(vf^2 - vi^2) = 2ad
Where:
- vf is the final velocity (which is 0 since the car needs to come to a stop)
- vi is the initial velocity
- a is the acceleration
- d is the distance traveled
a) To find the magnitude of acceleration when the car is traveling at 60 mi/h and needs to stop in a distance of 200 ft:
1. Convert the initial velocity to ft/s:
Initial velocity (vi) = 60 mi/h
1 mile = 5280 feet, 1 hour = 3600 seconds
vi = (60 mi/h) * (5280 ft/mi) / (3600 s/h) ≈ 88 ft/s
2. Plug the values into the equation of motion:
(0 - (88 ft/s)^2) = 2a * 200 ft
3. Simplify and solve for a:
7744 ft^2/s^2 = 2a * 200 ft
15488 ft^2/s^2 = 400a
a ≈ 38.7 ft/s^2
Therefore, the car needs to have an acceleration of approximately 38.7 ft/s^2 to come to a complete stop in a distance of 200 ft when traveling at 60 mi/h.
b) To find the acceleration required to stop in 200 ft when the car is traveling at 110 mi/h:
1. Convert the initial velocity to ft/s:
Initial velocity (vi) = 110 mi/h
vi = (110 mi/h) * (5280 ft/mi) / (3600 s/h) ≈ 161.3 ft/s
2. Plug the values into the equation of motion:
(0 - (161.3 ft/s)^2) = 2a * 200 ft
3. Simplify and solve for a:
25981.7 ft^2/s^2 = 2a * 200 ft
51963.4 ft^2/s^2 = 400a
a ≈ 129.9 ft/s^2
Therefore, the car needs to have an acceleration of approximately 129.9 ft/s^2 to come to a complete stop in a distance of 200 ft when traveling at 110 mi/h.