Benny is traveling at a speed of 150 mi/h.  What is his speed in feet per second?
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
756848
Bob was travelling at 9.5 meters/sec when he started his jump. To gain maximum range he needed to jump at a perfect 45 degree angle. If he had done so, his initial velocity vertically would have been 9.5 /sqrt(2) = 6.72 meters/sec. His maximum height would occur at time t where 6.72-gt = 0.\n9.78t = 6.72\nt = 6.72/9.78 = 0.687 secs\n\nAfter that he would fall for another 0.687 secs till he hit the ground for a total of 2(0.697) = 1.37 secs\n\nIn this time, he would have traveled (6.72)(1.37) meters horizontally or 9.23 meters.\n\nSo his jump of 8.90 meters was 0.33 meters short of perfection.\n\nNote that air resistance (which is pretty important in reality) would have the effect of making the maximum achievable jump shorter than 9.23 meters. So his jump was better than it appeared..
310980
about 1000 mph.\nSome say 1040 mph.
1047333
A 150lb. person on the planet Mars would weigh 38 lbs. on Mars.\nIf an average 150 lb. person was able to jump 3 feet high on Earth then the same person on Mars should be able to jump\n(150/38) * 3 feet = 11.84 feet high.  If he could jump that high on Earth he would have been drafted 1st by the NBA and be a star player, lol.
694954
Variables: 150 rev/min  2.2h = 140 min \n(a) Difference in velocity (150 to 0) over difference in time (0 minutes to 140)\ndelta v / delta t   (150-0)/(0-140) = -1.0714 rev/min/min\nAcceleration is negative since it is slowing down.\n(b) during the slowdown? one less than the answer to (c)?\n(c) (v not times delta t) / acceleration absolute value  (150*140)/|-1.0741|  =  19553.07 revs\n(d) and (e) email if you really want, but this stuff should give you a start. Hope this isn't due for a couple days tho.
53689
The combined speed of the two cyclists is 25k/hr\nThey will meet in 2 hours.\nThe bee is flying 20k/hr for 2 hrs so it will fly 40K.
750320
If it takes 6 secs to reach the max, it will take an additional 6 secs to reach your starting point.  In this case it is 2560.  Then it is a matter of finding the time from that point to going to the ground.\n1st lets correct the orginal equations (acceleration due to gravity is down/negative)\nh = -16t^2 + 192t + 2560\n2nd - set the equation to zero since we want to find the time to ground.\n0 = -16t^2 + 192t + 2560\n16t^2 - 192t = 2560\nt^2 - 12t = 160\n(t^2 - 12t + 36) - 36 = 160\n(t - 6)^2 = 196\nt - 6 = 14\nt = 20\nConclusion: 6 seconds to reach max ht of 3136, another 6 seconds to reach starting point of 2560 and another 8 seconds to hit the ground.
747360
Well gosh....  \n\nsince the only linear unit mentioned in the question is 18 ft/s  let's assume that the 60 is going to give us acceleration in ft/sec^2 \n\nSo, the position function for this is the second integral of the acceleration function, right? \n\nso...  the first anti-derivative of 60t is  30t^2   which gives the velocity, so at the end of the boost phase, our rocket is moving 30 * 3 * 3 = 270 ft/s \n\nand the second antiderivative is 10t^3  right?  \n\nSo, the position at time T is 10 * 3 * 3 * 3 = 270 feet.  (odd coincidence, but there you are...) \n\nNow, the rocket begins a SECOND acceleration by gravity of -32 ft/s negative 'cause we're accelerating DOWN now... ....   but this time the acceleration is constant,  so \n\nV = gt   and s = 1/2  a  t^2 so, we're slowing from 270 feet per second to zero at the top...  and  t = v/g = 270 / 32 = 8.44 seconds and the height we coast to is the 270 we had at the end of boost plus   1/2  *  32  * 8.44 * 8.44 = 1140 feet.  \n\nThe parachute is going to open after 14 seconds...   so, 14-8.44 is 5.56 seconds of free-fall (and we'll ignore air resistance again) so, we go from zero to something in 5.56 seconds....   and position is 1140 - ( 1/2  * g  * t^2 ) again...   so, the parachute opens at \n\n1140 - (1/2  * 32  * 5.56 * 5.56 )  = 645 feet. \n\nad the velocity is  gt = -32 * 5.56 = -178 feet per second...  \n\nNow, for the NEXT five seconds, we slow under the effect of the parachute   to  -18 ft/sec -- so,  we slow by 160 feet per second in five seconds, so the acceleration of the parachute is (oddly) 32 feet/second squared (huh, how odd is that? )  \n\nSo, the distance traveled during that phase is Vo t + 1/2 at^2  \n\nSo, we have (-160 * 5) + (1/2*32*5*5) =  -800 + 400 = 400 feet.  \n\nand from 400 feet up, we continue down at 18 ft/s for an additional  22 seconds...  \n\nand the rocket lands at t = 22 + 5 +  14 + 3  = 44 seconds \n\nAnd yes, I dropped a bunch of decimal points.  Your original data was only to two significant digits....
771460
Since the crate is moving at constant speed, acceleration = 0, net force is 0.\n\nApplied Force 150 N --> [41.5 kg mass] <-- Friction F = 150 N\n\nFor sliding friction, F(friction) = u N , where u (mu) is the coefficient, and N is the Normal force (imagine an arrow pointing down from the mass).\nN = mass x g
490704
Let s = speed of the boat\n\n60 / (s - 10) = 140 / (s + 10)\n140(s - 10) = 60(s + 10)\n140s - 1400 = 60s + 600\n80s - 1400 = 600\n80s = 2000\ns = 2000 / 80\ns = 25\n\ncheck:\n60 miles at 15 mph = 4 hours\n140 miles at 35 mph = 4 hours\n\nThe answer is 25 mph.
883093
speed = 55km/h = 275/18 m/s\ndistance 250 + 520 = 770m\ntime = distance/speed\n= 770*18/275 Seconds\n= 50.4 seconds
494819
MORE OR LESS 135 KM/H.
680534
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.
685470
the speed = distance/time
406060
Mythbusters equation: Elavator speed = 55mph; jump speed 2-3 mph. 60 - 3 = 57 mph = SPLAT!!!\n\nPS The myhtbusters had a 50 story building, and a special contraption that launched the test dummy at the second before impact.
947787
In this equation, you have a right triangle.  After 3 hours, the first car has traveled 150 km south and the 2nd car ahs traveled 210 km west.  You can find the between these two cars at 3 hours by using the Pythagorean Theorem since you have a right triangle.\n\nDistance = sqrt(150^2+210^2) = 258.07 km\n\nThe general form of the Pythagorean Theorem is:\n\na^2 + b^2 = c^2, where:\n\na = distance of car 1\nb = distance of car 2\nc = distance between cars.\n\nTake the derivative to find how the rates are related to one another.  Take the derivative with respect to time:\n\n2a(da/dt) + 2b(db/dt) = 2c(dc/dt)\n\nIn this problem, we are looking for dc/dt.  These are our known values:\n\na = 150 km\nb = 210 km\nc = 258.07 km\nda/dt = 50 km/hr\ndb/dt = 70km/hr\n\nPlugging these values for the equation above, solve for dc/dt:\n\ndc/dt = [2a(da/dt) + 2b(db/dt)]/[2c] ------->\n\n[(2*150*50) + (2*210*70)]/(2*258.07)\n\nTherefore, dc/dt = 86.02 km/hr.  Now change to km/sec:\n\n(86.02 km / 1 hr)(1 hr / 60 min)(1 min / 60 sec) = 0.024 km/sec\n\n-----------\n\nHope this helps
1071909
roughly 11 meters per second, or 24 miles an hour.\n;o)