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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=(2.70m/s3)t, where the +y-direction is upward.What is the speed of the rocket when it is 325 m above the surface of the earth?

Respuesta :

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[tex]a = 2.7t[/tex] 
[tex]v = \int\limits^t_0 {2.7t} \, dt = \frac{2.7}{2} t^2[/tex] 
[tex]x = \int\limits^t_0 {\frac{2.7}{2} t^2} \, dt = \frac{2.7}{6} t^3[/tex]

Solve for v with x = 325.

Explanation:

The vertical acceleration of the rocket is given by :

[tex]a_y=2.7t[/tex]

Acceleration is given by, [tex]a=\dfrac{dv}{dt}[/tex]

[tex]\dfrac{dv}{dt}=2.7t[/tex]

[tex]v=\int\limits{2.7\ t.dt}[/tex]  

[tex]v=\dfrac{2.7t^2}{2}[/tex]..............(1)

Velocity is given by, [tex]v=\dfrac{dx}{dt}[/tex]

[tex]\dfrac{dx}{dt}=\dfrac{2.7t^2}{2}[/tex]

[tex]x=\int\limits{\dfrac{2.7}{2}t^2}.dt[/tex]

[tex]x=\dfrac{2.7}{6}t^3[/tex]

From above equation, we can find the value of t at x = 325 m

[tex]325=\dfrac{2.7}{6}t^3[/tex]

t = 8.97 s

Now put t = 8.97 s in equation (1) as :

[tex]v=\dfrac{2.7(8.97)^2}{2}[/tex]

v = 108.62 m/s

So, the speed of the rocket when it is 325 meters above the surface of earth is 108.62 m/s. Hence, this is the required solution.

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