Answer
Please see the work below.
Work Step by Step
We can find the average speed as
$v=\frac{\frac{220}{3.6}}{2}$
$v=30.556\frac{m}{s}$
Now we can find the shortest runway as
$d=vt$
We plug in the known values to obtain:
$d=(30.556)(29)=890m$
Essential University Physics: Volume 1 (4th Edition)
by Wolfson, Richard
Published by
Pearson
ISBN 10:
0-134-98855-8
ISBN 13:
978-0-13498-855-9
Chapter 2 - Exercises and Problems - Page 32: 65
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