The distance a plane travels before takeoff is given by d = t^2, where d is in meters and t in seconds. If the plane becomes airborne at a speed of 234 km/hr, how long will it take for the plane to take off?
2007-04-12 05:25:36 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
d = t^2
d/dt = velocity = 2t
2t = 234 * 1000 m/km * 1 hr/60*60 secs = 65
t = 65/2 = 32.5 secs
how many times are you going to ask the same type of question before you learn the method to solve these?
2007-04-12 05:29:56 · answer #1 · answered by Grant d 4 · 2⤊ 0⤋
initial airborne speed of 234 km/hr = (234*10^3)/ (60*60) m/s = 65 m/s
displacement, d=t^2
velocity, d(d)/dt = 2t
at time T, velocity is 65.
Hence 2T = 65
solving, T = 65/2 = 32.5s
sub T=32.5 into original equation of d=t^2,
d=1056.25m
Therefore, a distance of 1056.25m is travelled in 32.5s just before the plane becomes airborne.
2007-04-12 12:33:27 · answer #2 · answered by wong boeing 1 · 0⤊ 0⤋
The speed is the derivative of distance in respect to time:
234 km/hr = 234 * 1000m / 3600sec = 234000/3600 m/sec =
= 2340/36 m/sec = 65 m/sec
Now, let's v be the speed in m/sec
v=d'(t) = 2t
put v=65
65 = 2t
t=32.5
2007-04-12 12:34:09 · answer #3 · answered by Amit Y 5 · 0⤊ 0⤋