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3 years ago

Suppose that the space shuttle Columbia accelerates at 14.0 m/s2 for 8.50 minutes after takeoff.

Physics

1 answer:

givi [52]3 years ago

5 0

Answer:

A. speed = 7.14 Km/s

B. distance = 1820.7 Km

Explanation:

Given that: a = 14.0 m/ $s^{2}$ , t = 8.50 minutes.

But,

t = 8.50 = 8.50 x 60

= 510 seconds

A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;

v = u + at

where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

u = 0

So that,

v = 14 x 510

= 7140 m/s

The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.

B. the distance traveled can be determined by applying second equation of motion.

s = ut + $\frac{1}{2}$ a $t^{2}$

where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.

u = 0

s = $\frac{1}{2}$ a $t^{2}$

= $\frac{1}{2}$ x 14 x $(510)^{2}$

= 7 x 260100

= 1820700 m

The distance that the shuttle has traveled during the given time is 1820.7 Km.

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4 years ago

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Answer:

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