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Kitty [74]
3 years ago
6

An Airbus A350 is initially moving down the runway at 6.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the

engines give the plane a constant acceleration of 1.8 m/s^2. The plane then travels a distance of 1500 m down the runway before lifting off. How long does it take from the application of the acceleration until the plane lifts off, becoming airborne?
Physics
1 answer:
alexdok [17]3 years ago
8 0

Answer:

t=67.7s

Explanation:

From this question we know that:

Vo = 6m/s

a = 1.8 m/s2

D = 1500m

And we also know that:

X=V_{o}*t + \frac{a*t^{2}}{2}   Replacing the known values:

1500=6t+0.9*t^{2}    Solving for t we get 2 possible answers:

t1 = -44.3s   and t2 = 67.7s    Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:

t = 67.7s

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We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

a=\frac{v-u}{t}

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We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

a=\frac{v-u}{t}

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v  is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

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The change in velocity during the 1-s interval is given by

\Delta v = v -u

where here we have

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u = 0 is the initial velocity (at the highest point)

Substituting, we find

\Delta v = -9.8 m/s - 0=-9.8 m/s

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The change in velocity during the overall 2-s interval is given by

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m is the mass of the ball

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mg = ma

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