A space vehicle accelerates uniformly from 65m/s at t=0 to 162m/s at t=10.0s. How far did it move between t=2.0s and t=6.0s?
2007-09-21 02:01:59 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
395.8miles
2007-09-21 02:19:25 · answer #1 · answered by Zuker 5 · 0⤊ 2⤋
v = u + at is the formula to be used for calculating acceleration a from the initial velocity u, the final velocity v and the time t.
162 = 65 + a.10
10a = 162 - 65 = 97
a = 9.7 m/sec^2
s = ut + 1/2 at^2 is the formula to be used for calculating the distance covered.
s1 = 65x2 + 1/2 x 9.7 x 2^2
s2 = 65 x 6 + 1/2 x 9.7 x 6^2
s2 - s1 is the distance moved between 2 and 6 seconds.
= 65 x (6 - 2) + 9.7/2 (6^2 - 2^2)
= 65 x 4 + 4.85 x (36 - 4)
= 260 + 4.85 x 32
= 260 + 155.2
= 415.2 m
That is the distance moved by the space vehicle.
2007-09-21 02:55:27 · answer #2 · answered by Swamy 7 · 4⤊ 0⤋
use v=vo + at, v= 162 vo=65 t=10 to find a, then use change in x = x - xo = vot+ 1/2at^2, v0=? t=(6-2).. to use vo at time 2 not 0.. vo at time 2s = (vo at time 0)+a(2).. the answer is (x-xo)
2007-09-21 02:26:25 · answer #3 · answered by Anonymous · 2⤊ 0⤋
acceleration = delta v / delta t = (vf - vi) / (tf - ti)
position (t) = v0t + 1/2 at^2
= v0t + 1/2 t^2 (vf - vi) / (tf - ti)
Find position at the first time and at the second time. Take the difference.
2007-09-21 02:25:05 · answer #4 · answered by Anonymous · 0⤊ 0⤋