Suppose that the space shuttle Columbia accelerates at 14.0 m/s2 for 8.50 minutes after takeoff. A. What will be the speed of the shuttle at the end of 8.50 minutes?
B. How far will the shuttle have traveled during this time?

A. speed = 7.14 Km/s

B. distance = 1820.7 Km

Explanation:

Given that: a = 14.0 m/, 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 + a

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

u = 0

s = a

  = x 14 x

 = 7 x 260100

 = 1820700 m

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

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