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A spacecraft is traveling with a velocity of v0x = 6580 m/s along the +x direction. Two engines are turned on for a time of 907 s. One engine gives the spacecraft an acceleration in the +x direction of ax = 1.81 m/s2, while the other gives it an acceleration in the +y direction of ay = 8.05 m/s2. At the end of the firing, what is a) vx and b) vy?

Answer :

amhugueth

Answer:

[tex]v_{x}=8222m/s\\v_{y}=7301m/s[/tex]

Explanation:

From the exercise we know that

[tex]v_{ox}=6580m/s\\a_{x}=1.81m/s^2\\a_{y}=8.05m/s^2[/tex]

[tex]t=907s[/tex]

Knowing the following formula we can calculate the final velocity

[tex]v=v_{o}+at[/tex]

[tex]v_{x}=v_{ox}+a_{x}t[/tex]

[tex]v_{x}=6580m/s+(1.81m/s^2)(907s)=8222m/s[/tex]

[tex]v_{y}=v_{oy}+a_{y}t[/tex]

[tex]v_{y}=(8.05m/s^2)(907s)=7301m/s[/tex]

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