Question

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.

Answer 1

Answer:

Explanation:

Given

acceleration in x direction

$a_x=5.10 m/s^2$

Final velocity in x direction $v_x=3500 m/s$

time taken=150 s

let initial velocity in x direction be $u_x$

$v_x=u_x+a_x\times t$

$3500-5.10\times 150=u_x$

$u_x=2735 m/s$

Now solve for Y direction

$v_y=u_y+a_y\times t$

$4046=u_y+7.3\times 150$

$u_y=4046-1095=2951 m/s$

$tan\theta =\frac{u_y}{u-x}=\frac{2951}{2735}$

$tan\theta =1.078$

$\theta =47.14 ^{\circ}$ ACW from x-axis