1 )A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.0 km, the jet is moving with a speed of 400 m/s. What is the jet's acceleration, assuming it to be a constant acceleration?
2)A plane accelerates from rest at a constant rate of 5.00 along a runway that is 1800 long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time needed to take off?
2007-09-22 10:31:52 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
1) vf^2 - vi^2 = 2ax
a = (vf^2 - vi^2) / 2x
They give you the speeds and the distance. Plugnchug.
2) For a plane accelerating from rest, the position is:
d = 1/2 at^2
t = sqrt (2d/a)
They give you the distance and a. Plugnchug.
They give you
2007-09-22 10:38:51 · answer #1 · answered by Anonymous · 0⤊ 0⤋
1. V(final)^2 - V(initial)^2 = 2ad
400^2 - 300^2 = 2 * 4000 a (It's 4000m instead of 4.0 km for the distance so that we can keep the velocities in m/s)
70,000 / 8,000 = a = 8.75 m/s^2
2. I'm assuming the acceration is in m/s^2 and the length of the runway is 1800 m. Units are VERY important when doing this type of calculation. Please include them next time.
d = v*t +1/2 * a * t^2
"at rest" ~> V = 0
d = 1/2 * a * t^2
1800 = 2.5 t^2
720 = t^2
t = 26.8 s
2007-09-22 17:36:29 · answer #2 · answered by lhvinny 7 · 0⤊ 1⤋
2)e=1/2*5 t^2 so 1800 *2/5=t^2 so t=26.8
I hope all information is given in coherent units
2007-09-22 17:44:03 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋
1) 400m/s - 300 m/s divided by 4.0
2) No idea
2007-09-22 17:37:08 · answer #4 · answered by penguingirl93 1 · 0⤊ 1⤋