A rocket starts from rest and moves upward from the surface of the earth. For the first 10s of its motion, the vertical acceleration of the rocket is given by ay=(3.00m/s3)t, where the +y-direction is upward. What is the height of the rocket above the surface of the earth at t = 10s ? What is the speed of the rocket when it is 350m above the surface of the earth? a) I tried s=1/2*at^2 = ((3.00 m/s^3)(10^2 s)) / 2 = 150m but that was not the correct answer. I am also not sure how the seconds cancel out (my equation really gives me m/s rather than just m). b) To find velocity at 350m distance above the earth, I have to find time so 350m = (1/2)(3.00 m/s^3)t^2 so t=15.28s Then v=at as the initial velocity is zero, so v = (3.00 m/s^3)(5.28s) = 45.84 m/s Can someone please let me know what I'm doing wrong?

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day