A jet accelerates from 15.98 m/s to 34.05 m/s in 12.90 seconds. The displacement (to the nearest meter) of the jet during this time interval is:
2007-09-18 13:50:09 · 2 answers · asked by ags101 2 in Science & Mathematics ➔ Physics
A simpler formula is
v(ave) = (v0+v1)/2 = 25.015 m/s
x = v(ave)t = 322.694 m
2007-09-19 16:34:54 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
ok so first you find accerlation using this formula:
v = v0 + at
34.05 = 15.98 + a(12.90)
34.05 - 15.98 = 12.90a
18.07 = 12.90a
a = 1.4 m/s^2
Now use this in the following equation:
v^2= v0^2 + 2a(deltaX)
(34.05)^2 = (15.98)^2 + 2(1.4)(x)
1159.4 = 255.36 + 2.8x
1159.4 - 255.36 = 2.8x
904.04 = 2.8x
x= 322.87
by the way, you can also use the equation
(delta x) = v0t + 1/2(a)(t^2)
x = (15.98 * 12.90) + 1/2(1.4)(12.90^2)
=206.142 + 116.487
=322.629
so it would be around 323 meters
2007-09-18 21:06:32 · answer #2 · answered by redblitz528 1 · 0⤊ 0⤋