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 645 s, at which point the spacecraft has velocity components of vx = 3780 m/s and vy = 4470 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:

v=545.41 $\frac{m}{s}$

β=-25.93

Explanation:

Give the acceleration in 'x' and 'y' also the time can find the initial velocity using equation of uniform acceleration motion

For axis x

$a_{x}=5.10\frac{m}{s^{2}} \\v_{fx}=3780\frac{m}{s} \\v_{fx}=v_{ox}+a_{x}*t\\v_{ox}=v_{fx}-a_{x}*t\\v_{ox}=3780 \frac{m}{s}-5.1\frac{m}{s^{2} }*645s\\v_{ox}=490.5\frac{m}{s}$

For axis y

$a_{y}=7.30\frac{m}{s^{2}} \\v_{fy}=4470\frac{m}{s} \\v_{fy}=v_{oy}+a_{y}*t\\v_{oy}=v_{fy}-a_{y}*t\\v_{oy}=4470\frac{m}{s}-7.30\frac{m}{s^{2} }*645s\\v_{oy}=-238.5\frac{m}{s}$

Maginuted

$v=\sqrt{v_{xo}^{2} +v_{yo}^{2} } \\v=\sqrt{490.5x^{2} +(-238.5)^{2} }\\v=545.41 \frac{m}{s}$

The  direction is knowing when find the angle so

$\beta =tan^{-1}*\frac{v_{yo} }{v_{xo}}\\\beta =tan^{-1}*\frac{-238.5}{490.5}\\\beta =tan^{-1}*-0.48\\\beta =-25.93$