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A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of ax = 5.10 m/s2, while the one on the back gives an acceleration component in the y direction of ay = 7.30 m/s2. The engines turn off after firing for 150 s, at which point the spacecraft has velocity components of vx = 3500 m/s and vy = 4046 m/s. What was the magnitude and the direction of the spacecraft's initial velocity before the engines were turned on? Express the magnitude as m/s and the direction as an angle measured counterclockwise from the +x axis.

Respuesta :

Answer:

Explanation:

Given

acceleration in x direction

[tex]a_x=5.10 m/s^2[/tex]

Final velocity in x direction [tex]v_x=3500 m/s[/tex]

time taken=150 s

let initial velocity in x direction be [tex]u_x[/tex]

[tex]v_x=u_x+a_x\times t[/tex]

[tex]3500-5.10\times 150=u_x[/tex]

[tex]u_x=2735 m/s[/tex]

Now solve for Y direction

[tex]v_y=u_y+a_y\times t[/tex]

[tex]4046=u_y+7.3\times 150[/tex]

[tex]u_y=4046-1095=2951 m/s[/tex]

[tex]tan\theta =\frac{u_y}{u-x}=\frac{2951}{2735}[/tex]

[tex]tan\theta =1.078[/tex]

[tex]\theta =47.14 ^{\circ}[/tex] ACW from x-axis

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