A spaceship moving with an initial velocity of 58.0 meters/second experiences a uniform acceleration and attains a final velocity of 153 meters/second. What distance has the spaceship covered after 12.0 seconds?

Answer 1

d = V0 * t + 1/2 a t ** 2
where
average acceleration a = (V1 - V0) / dt

V0 = 58 m/s
V1 = 153 m/s
dt = 12 s
a = (153-58)/12 m/(s*s)
a = 7.9 m/(s*s)
d = 58 * 12 m/s *s + .5 * 7.9 * (12 * 12) m/(s*s) * (s*s)
d = 696 + 1140 m
d = 1836 m

Answer 2

P.S.
I forgot to multiply by the .5 in the second to last step .
It should read :
d = 696 + .5 * 1140 m
d = 1266 m

Answer 3

100

Answer 4

lad