160 feet in 5min 20 seconds?
411654
exactly 2 miles per hour.
100012
A would take 22 minutes to get to 15 miles and 32 minutes to get to 20 miles. Unfortunately, there's insufficient information to be precise on B's journey. Did he spin off before the start?.. or at which point en route? Guessing it, I would say A arrived at 15 miles and at the 20 miles point first. NB. I have had to make certain assumptions in your question to arrive at this answer (eg I asumed A slows down to 40mph after 4 miles?)
152036
13,200,000 Milliseconds are in 220 minutes \n\n220 x 60 x 1000
749848
The maximum speed is about 220-235 MPH, and it can be achieved in about 20 seconds of freefall, in a verticle dive (hands flat against the legs, body very straight, head down attitude),  which would require only about 6000 feet of altitude. Then the parachutist would have to assume the 'flat and stable' body position, after about 25 seconds,  in which the back is arched, arms outspread, hands flat, facing the earth, legs slightly spread and slightly bent. This will allow him/her to slow down to the normal 120 MPH speed for the opening of the parachute at a safe altitude.  To open a parachute at the higher speeds could easily cause sturctural failure of both the parachute,  and the human body.  In civilian skydiving, 2500 feet is considered the minimum safe altitude for parachute deployment. Only licenced Experts are allowed to go lower,  in the US the minimum for the "D" (Expert) licencee is 2000 feet. In Europe it is 600 meters, or 1960 feet.  Many military drops are conducted at lower altitudes,  but rarely with freefall rigs. The openings on these parachutes are automatic and happen immediately as the jumper leaves the aircraft. Federal Aviation Regulations do not regulate military operations.
882711
a)\nRate = Distance/time = (24000 ft)/(20 min) = 1200 ft/min\n\nb)\nTo write A(t), you know that the max. value is 24000 and the plane descends at a constant, linear rate of 1200 ft/min.  Thus, you can write a linear equation of this model:\n\nA(t) = 24000 - 1200t, where t = time (in minutes)\n\nPlugging in 0 (at initial time), the height is 24000 feet, which matches the fact in the problem.\n\nc)\nThough linear functions have infinite ranges and domains, this problem is limited to the realism of the problem.  The range is only limited from the maximium value (24000 ft) to the ground (0 ft).  You can't have a negative value for altitude because that would not make sense in the problem.  The domain is limited to the time in air (20 min).  You can't have a negative time and time after 20 min is irrelevant because the plane has already landed.  Summarizing:\n\nDomain: 0 <= t <= 20\nRange: 0 <= A(t) <= 24000\n\n--------\n\nHope this helps
714444
About 175km \n\nabout 2hours 29 minutes
299762
If you can run 20 seconds without problems....you are not fat
554079
(150mi/1h)*(1h/60min)*(1min/60sec)*(5280ft/1mile)=220ft/sec\n\nFirst, convert miles to feet: 150 miles * 5280 ft/mile = 792,000.  This is your feet/hour.\n\nNow, convert hours to seconds: 1 hr * 60 min/hr * 60 sec/min = 3600 seconds in one hour.\n\nDivide your feet/hour by seconds/hour:\n\n792,000/3600 = 220 ft/sec
63833
50 feet or about 2-3 seconds
441442
you have your units wrong, do you mean 5m/s or 5m/s^2 and you mean 20m from the ground?\nIf acc=5 when height=20 speed= 14.14 in upwards direction (using 2ax=v^2-u^2)\nIf initial speed is therefore -14.14m/s then at t=1 speed=-4.34m/s so it is still going upwards\nt=2 speed= 5.45m/s it is now going down. Both last results worked out using v=u + at\n(Sorry I assumed you meant 5m/s^2 straight up, you need the angle at which it accelerates aswell)
933645
Would it help if you knew that there are 3600 seconds in an hour, and 5280 feet in a mile? D=R*T
1148347
One billionth of a second.
362084
45 mins plus extra time
549547
one second is too long
924813
(288^2)/(2*32.2) is the maximum height.\nat time 288/32.2 seconds\nit hits ground at time time   t+576/32.2, where t is obtained from\n45=288+.5*32.2*t^2
925006
Since the magnitude of the acceleration is v^2/r, and the radius of the Earth is approximately 4000 miles (21120000 ft), you would have to solve the following equation: v^2/21120000=32\nv^2=675840000\nv=25997feet per second  or\nv=4.92 miles per second
112380
Try timing your strokes per minute then multiply by 60.