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At launch a rocket ship weighs 4.5 million pounds. When it
Chapter 2, Problem 29E(choose chapter or problem)
Launch of the Space Shuttle. At launch the space shuttle weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 km/h, and at the end of the first 1.00 min its speed is 1610 km/h.
(a) What is the average acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\)) of the shuttle
(i) during the first 8.00 s, and
(ii) between 8.00 s and the end of the first 1.00 min?
(b) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the shuttle travel
(i) during the first 8.00 s, and
(ii) during the interval from 8.00 s to 1.00 min?
Questions & Answers
QUESTION:
Launch of the Space Shuttle. At launch the space shuttle weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 km/h, and at the end of the first 1.00 min its speed is 1610 km/h.
(a) What is the average acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\)) of the shuttle
(i) during the first 8.00 s, and
(ii) between 8.00 s and the end of the first 1.00 min?
(b) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the shuttle travel
(i) during the first 8.00 s, and
(ii) during the interval from 8.00 s to 1.00 min?
ANSWER:Step 1 of 3
Given data:
The mass of the space shuttle is m = 4.5 million pounds.
The initial velocity is u = 0 m/s.
The final velocity is v = 161 km/h in time t = 8.00 s.
The final velocity is V = 1610 km/h in t = 1.00 min.
Let us first convert the given velocities into m/s units.
We have,
\(\frac{{1\;{\rm{km}}}}{{1\;{\rm{h}}}} = \frac{{1000\,{\rm{m}}}}{{3600\;{\rm{s}}}}\)
\(1\;{\rm{km/h}} = \frac{5}{{18}}\,{\rm{m/s}}\)
Therefore,
\(161\;{\rm{km/h}} = \frac{5}{{18}} \times 161\;{\rm{m/s}}\)
\(161\;{\rm{km/h}} = 42.72\,{\rm{m/s}}\)
\(1610\;{\rm{km/h}} = \frac{5}{{18}} \times 1610\;{\rm{m/s}}\)
\(1610\;{\rm{km/h}} = 427.2\,{\rm{m/s}}\)