Find the velocity of the object after one second.
v = vo + at
v = (0 m/s) + (9.8 m/s^2)(1 s)
v = 9.8 m/s
Now, using that, you can find the displacement in that one second between 1 and 2.
d = vot + (1/2)at^2
d = (9.8 m/s)(1 s) + (1/2)(9.8 m/s^2)(1 s)^2
d = 14.7 m
The guys before me was almost right. But instead of it being vertical it is horizontal.
Answer:
v = 6t² + t + 2, s = 2t³ + ½ t² + 2t
59 m/s, 64.5 m
Explanation:
a = 12t + 1
v = ∫ a dt
v = 6t² + t + C
At t = 0, v = 2.
2 = 6(0)² + (0) + C
2 = C
Therefore, v = 6t² + t + 2.
s = ∫ v dt
s = 2t³ + ½ t² + 2t + C
At t = 0, s = 0.
0 = 2(0)³ + ½ (0)² + 2(0) + C
0 = C
Therefore, s = 2t³ + ½ t² + 2t.
At t = 3:
v = 6(3)² + (3) + 2 = 59
s = 2(3)³ + ½ (3)² + 2(3) = 64.5
Answer:
For starters, “dark side” is a misnomer. The far side may be constantly facing away from Earth, but as it orbits the Earth it sometimes faces the sun just as the bright side does. Thus the far side is subject to the same day/night temperature extremes as the side facing Earth.
Explanation:
