Arocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical acceleration of the rocket is given by ay=(3.00m/s3)t, where the +y-direction is upward. part a what is the height of the rocket above the surface of the earth at t = 10.0 s ?

Hello

1) The total vertical acceleration of the rocket is , where and  is the gravitational acceleration that points downward.

2) This is an accelerated motion, so we can write the law of motion as

where S is the total distance covered by the rocket after time t. We can then substitute into the previous equation and we get

And so, after  s, we get  m, which is the height reached by the rocket after 10 seconds.

The upward acceleration is
a(t) = (3.00 m/s³)(t s) = 3t m/s²

Because the rocket starts from rest, the upward velocity at time t is


The distance traveled in 10 sec is


500 m