How does (X-Xo) give the min length needed for the run way, I understand how to

Question

How does (X-Xo) give the min length needed for the run way, I understand how to get the answer (193m) but I don't understand why we assume that (X-Xo) gives you the length you're looking for

Runway design. You are designing an airport for small EXAMPLE 2-9 planes. One kind of airplane that might use this airfield must reach a speed before takeoff of at least 27.8 m/s (100 km/h), and can accelerate at 2.00 m/s2. (a) If the runway is 150 m long, can this airplane reach the required speed for takeoff? (b) If not, what minimum length must the runway have? APPROACH The plane's acceleration is constant, so we can use the kinematic equations for constant acceleration. In (a), we want to find v, and we are given Known Wanted x 150 m a = 2,00 m/s SOLUTION (a) Of the above four equations, Eq. 2-12c will give us v when we know vo, a, x, and xo: 0 + 2(2.00 m/s2)(150 m) 600 m2/s7 24.5 m/s 600 m2/s2 = This runway length is not sufficient. (b) Now we want to find the minimum length of runway, x - xo, given v 27.8 m/s and a -2.00 m/s2. So we again use Eq. 2-12c, but rewritten as (27.8 m/s)2-0 2(2.00 m/s2) U2- x-xo) = 193 m A 200-m runway is more appropriate for this plane. NOTE We did this Example as if the plane were a particle, so we round off our answer to 200 m

Explanation / Answer

Since In the question given that minimum speed required to take-off is 27.8 m/sec

Now it means that plane cannot take off until it reaches the speed of 27.8 m/sec, Since initial speed is 0 m/sec and acceleration of plane is constant and equal to 2.00 m/sec^2. So

Now the distance required to reach this speed with given acceleration will be minimum length of runway.

it means the plane will start at one border point of runway with 0 m/sec velocity and it will reach this minimum required velocity if runway has length of 193 m (200 m).

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