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A rocket 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 ?

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skyluke89
Hello

1) The total vertical acceleration of the rocket is [tex]a=a_y-g[/tex], where [tex]a_y=3~m/s^3 t[/tex] and [tex]g=9.8~m/s^2[/tex] is the gravitational acceleration that points downward.

2) This is an accelerated motion, so we can write the law of motion as
[tex]S= \frac{1}{2}at^2 [/tex]
where S is the total distance covered by the rocket after time t. We can then substitute [tex]a=a_y-g[/tex] into the previous equation and we get
[tex]S= \frac{1}{2}(a_y-g)t^2 = \frac{1}{2}(3t-9.8)t^2[/tex]
And so, after [tex]t=10.0[/tex] s, we get [tex]S=1010[/tex] 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
[tex]v(t) = \int_{0}^{t} a(t)dt = \int_{0}^{t} 3t \,dt = \frac{3}{2} t^{2}[/tex]

The distance traveled in 10 sec is
[tex]y = \int_{0}^{10} v(t)dt = \frac{3}{2} \int_{0}^{10}t^{2} dt = \frac{1}{2} [t^{3}]_{0}^{10} = \frac{1}{2}(1000) = 500 \, m [/tex]

Answer: 500 m

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