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A cough travels how many mph?
Coughing is an action the body takes to get rid of substances irritating the air passages, and a cough is usually started to clear a buildup of phlegm (mucus) in the trachea. When a person coughs, air may move through this passage at up to 300 MPH. Source: http://www.nativeremedies.com/ailment/persistent-coughing-info.html
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132 feet per second at 90 mph.
time= distance/rate =318miles / 50mph =6.36 hours
85 Miles per Hour = 124.67 Feet per Second
80 and 2/3 feet per second or 80'8" per second
At 50 mph, a car travels 73.33 feet every second.
125 mph = 183.33 feet per second.
At 30 mph, an object is moving at about 44 feet every second.
To convert mph to feet per second, always multiply mph by 22/15. As an extension, to convert mph to feet per minute, always multiply mph by 88. To answer your question, (30) x… (22/15) = 44 feet per second
41.54 minutes (rounded)
About 12.75 miles at that rate of travel.
At 100 mph, an object is traveling about 146.67 feet every second. Therefore, to travel 60 feet at 100 mph would require about 0.41 seconds.
75 mi/hr = 110 ft/sec
It depends on several factors. Bullet can travel as slow as a few hundred feet per second up to several thousand feet per second. 1000 feet per second is about 682 MPH.
That depends upon for how long it is moving: Every day it will travel 50 miles/hr x 5280 ft/mile x 24 hr/day = 6336000 ft/day Every hour it will travel 6336000 ft/day ÷ 2…4 hr/day = 264000 ft/hr Every minute it will travel 264000 ft/hr ÷ 60 min/hr = 4400 ft/min Every second it will travel 4400 ft/min ÷ 60 sec/min = 731/3 ft/sec ≈ 73.33 ft/sec * Added - The simple formula that should be used is converting MPH to feet per second (FPS). This is done by multiplying by 1.467, or for an easy rough estimate, 1.5. This result in 50mph = 73.35 fps. This is the method most often used by engineers and accident investigators. This can be used for determining braking distances for a car as well, once the deceleration rate is identified.
That depends on the deceleration applied, in which case time = 2x45 mph/a = 90/a with a (the deceleration) measured in miles per hour per hour to give the time in hours. … If you mean the time that would be taken using the stopping distances in the UK Highway Code, then, assuming a constant deceleration (from the moment the brakes are applied after the thinking distance): The stopping distance at 45 mph is 45 ft (thinking distance) + 101.25 ft (braking distance) [braking distance is speed2 / 20 ft = 452 / 20 ft = 101.25 ft] The thinking distance is travelled at 45 mph, giving: think_time = 45 ft / 45 mph = 45 ft /(45 x 5280 ft / 3600 seconds) = 45 x 3600 / (45 x 5280) seconds = 3600/5280 seconds = 15/22 seconds (This time is constant for all the emergency stopping distances given in the Highway Code.) Two equations of motion can be used to find the stopping time knowing the initial speed and distance (the final speed is zero): final_velocity2 = initial_velocity2 + 2 x acceleration x distance final_velocity = 0 → acceleration = - initial_velocity2 / (2 x distance) distance = initial_velocity x time + 1/2 x acceleration x time2 → distance = initial_velocity x time - (1/4 x initial_velocity2 / distance) x time2 → time2 - (4 x distance / initial_velocity) x time + 4 x distance2 / initial_velocity2 = 0 → (time - 2 x distance/initial_velocity)2 = 0 → breaking_time = 2 x distance / initial_velocity = 2 x (101.25 ft) / (45 x 5280 / 3600 ft per sec) = 202.5 x 3600 / (45 x 5280) seconds = 9x15/2x22 → total_stopping_time = 15/22 + 9x15/2x22 seconds = 11x15/2x22 seconds = 15/4 seconds = 33/4 seconds = 3.75 seconds.