```
CATAGORY: Mechanics
QUESTION: A Boeing 747 Jumbo Jet has a length of 57m. The
runway on which the plane lands intersect another runway. The width of the
intersection is 25.0m.  The plane decellerates through the intersection at
a  rate of 5.70 m/s^2 and xlears it with a final speed of 45.0m/s.  How
much time is needed for the plane to clear the intersection.
```

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```From the middle figure below we see that the front part of the 747 has
crossed the intersection and there is 25 more meters to go before the
tail crosses over the intersection.  The plane is regarded as being in the
intersection so long as any part of the plane extends into the space of
the intersction.  For the nose of the plane to get to the point it is now
it traveled a distance of 57 m from when it was just touching the left
side of the intersection

Before

____
____/   |\_
============\
=====>______/
\____|>   \
UU     OO
---------|==========|----------------|-----
0m          25m             57m

Intersection
During
---                 ____
\  \_______________/   |\_
<==========================\
-----_____<=====>______/
\____|>   \
UU     OO
---------|==========|----------------|-----
0m          25m             57m

Intersection

After

---                 ____
\  \_______________/   |\_
<==========================\
-----_____<=====>______/
\____|>   \
UU     OO
---------|==========|----------------|-----
0m          25m             57m

Intersection

The tail has to now travel  adistance of 25 m to clear the
intersection which during this time the nose also travels forward another
25 meters.the plane has thus traveled an total of 57 m +25 m = 82m

The equation which relates velocity to acceleration and displacement is

Where d is the displacement, v and vo are the final and initial
velocities and a is the acceleration. Solving (1) for vo gives

(2)   vo^2 = v^2 - 2ad

so

substitution of v = 45 m/s, a = -5.7m/s^2 and d = 82 m gives

(4)  vo =sqrt((45 m/s)^2 - 2(-5.7m/s^2)82 m) = 54.4 m/s

Next displacement is related to time by

(4)  d = do + vo t + (1/2)a t^2

Here do =0m, d =82m, vo=54.4m, and a=-5.7m so

(5)  82m = (54.4m/s)t - (2.85m/s^2) t^2

or

(6)   (2.85m/s^2) t^2 - (54.4m/s)t +82m = 0m

(7) 2.85 t^2 - 54.4m/s t +82 = 0

gives

(8)       -(-54.4) +/-sqrt((-54.4)^2 - 4(2.85)82)
t =  ----------------------------------------
2(2.85)

54.4 +/-sqrt(2959.36 - 934.8)
= -------------------------------
5.7

54.4 +/-45.0
= ---------------
5.7

=  (17.5, 1.6)

We take the smaller value because the plane does not cross the
intersection, stop and then reverse its course and cross again! Only the
first crossing counts.  Thus

************
*          *
* t = 1.6s *
*          *
************
```

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