
#1
January 30th, 2013,
14:04
A light plane must reach a speed of 33 m/s for takeoff. How long must the runway be if the plane has a constant acceleration of 3.8 m/s^{2} ?
so, I am new to kinomatic equations. We are using the 4 Basic ones.
need to see this one worked out if I may.
Thanks

January 30th, 2013 14:04
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#2
January 30th, 2013,
14:05
What is the target variable? That is, what is it for which you're trying to solve?

#3
January 30th, 2013,
14:29
Thread Author

Pessimist Singularitarian
#4
January 30th, 2013,
14:33
What is it that you are given? And how may we relate these values?

#5
January 30th, 2013,
14:49
Another way of saying what MarkFL said is this: can you write down a relevant, correct equation involving the target variable of distance?

#6
March 10th, 2013,
15:31
Originally Posted by
mathkid3
A light plane must reach a speed of 33 m/s for takeoff. How long must the runway be if the plane has a constant acceleration of 3.8 m/s^{2} ?
so, I am new to kinomatic equations. We are using the 4 Basic ones.
need to see this one worked out if I may.
Thanks
I think the OP meant the kinematic equations, 4 basic ones meaning:
1) x=x_{0}+V_{0}t+1/2at^{2}
2) V=V_{0}+at
3) a=constant
4) V^{2}=V_{0}^{2}+2a(xx_{0})
If this is the case, the answer is a matter of determining starting values.
We can assume that the plane starts off at point 0 on the runway meaning that x_{0}=0. we can also assume that the plane's initial velocity is 0, V_{0}=0 (before moving). now we are given the acceleration, a=3.8m/s^{2}, and final velocity, V=33m/s. From this we can find how long it will take the plane to reach the final velocity, using equation 2:
33m/s=0+(3.8m/s^{2})t
t=(33/3.8)s
Now that we have a value for t, we can use equation 1 to find the total runway length, x, required to reach takeoff speed:
x=0+0(t)+1/2at^{2}
=(1/2)(3.8m/s^{2})((33/3.8)s)^{2}
=143.30m
therefore the minimum amount of runway needed to reach the plane's necessary takeoff speed is 143.30 meters.