This problems a perfect application for this acceleration formula:
Distance = (1/2) (acceleration) (time)² .
During the speeding-up half: 1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half: 1,600 meters = (1/2) (1.3 m/s²) T²
Pick either half, and divide each side by 0.65 m/s²:
T² = (1600 m) / (0.65 m/s²)
T = square root of (1600 / 0.65) seconds
Time for the total trip between the stations is double that time.
T = 2 √(1600/0.65) = <em>99.2 seconds</em> (rounded)
Answer: Try C
Explanation:
It's the only one that makes since.
The first person to say the Earth orbited the sun was Nicolaus Copernicus
Answer:
The distance traveled during its acceleration, d = 214.38 m
Explanation:
Given,
The object's acceleration, a = -6.8 m/s²
The initial speed of the object, u = 54 m/s
The final speed of the object, v = 0
The acceleration of the object is given by the formula,
a = (v - u) / t m/s²
∴ t = (v - u) / a
= (0 - 54) / (-6.8)
= 7.94 s
The average velocity of the object,
V = (54 + 0)/2
= 27 m/s
The displacement of the object,
d = V x t meter
= 27 x 7.94
= 214.38 m
Hence, the distance the object traveled during that acceleration is, a = 214.38 m
