<p>An airplane taking off from a runwany travels 3600 ft before lifting off. If it starts from rest, moves woth constant acceleration, and makes the run in 30 seconds, with what speed does it lift off??</p>
<p>my way: 3600ft/30sec= 120 ft/sec, however the answer seems to be incorrect( what am i doing wrong)</p>
<p>It's acceleration. Meaning, that the speed it starts will not be the same speed when it lifts off. It is upsloping. </p>
<p>So when t=0, x =0
And when t=30, x=3600</p>
<p>We have to figure out the equation...I might be totally wrong but</p>
<p>a(t)=3600 (i think???)
so v(t)=3600x + C</p>
<p>ummm, I dont know what to do now, without help, so Ill stop and hope someone does.</p>
<p>that's close..</p>
<p>you know
v(0) = 0
s(0) = 0
s(30)=3600</p>
<p>and you know
a(t) = C
therefore
v(t) = Ct + D, so v(0) = 0(C) + D = 0, so D = 0.
so v(t) = Ct
and then:
s(t) = 0.5Ct^2 + E, and s(0) = 0(C) + E = 0, so E = 0.
so s(t) = 0.5Ct^2.</p>
<p>you also know that s(30) = 3600, so you can solve out for C, which should be 8.</p>
<p>so then to find out the velocity after 30 seconds, just plug that back in -- so v(30) = 240 ft/sec.</p>
<p>that should be right..</p>
<p>This is like honors physics problems...</p>
<p>D=delta symbol</p>
<p>DX=0.5(Vfinal+Vinitial)t</p>
<p>3600=0.5(Vfinal)30</p>
<p>3600=15Vfinal</p>
<p>Vfinal=240 ft/sec</p>
<p>But you guys are in calculus...so you have to make the equation yourself. I remember this from last year in physics.</p>
<p>thanks indigo. kinda sorta not really was on the right track =)</p>
<p>i really appreciate all your help.</p>