Answer:
A. speed = 7.14 Km/s
B. distance = 1820.7 Km
Explanation:
Given that: a = 14.0 m/[tex]s^{2}[/tex], 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 + [tex]\frac{1}{2}[/tex]a[tex]t^{2}[/tex]
where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.
u = 0
s = [tex]\frac{1}{2}[/tex]a[tex]t^{2}[/tex]
= [tex]\frac{1}{2}[/tex] x 14 x [tex](510)^{2}[/tex]
= 7 x 260100
= 1820700 m
The distance that the shuttle has traveled during the given time is 1820.7 Km.
