Asked by luisa
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?
Answers
Answered by
john
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
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
Answered by
john
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
I forgot to multiply by the .5 in the second to last step .
It should read :
d = 696 + .5 * 1140 m
d = 1266 m
Answered by
kiara
100
Answered by
e
lad
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.