Answer:
a)24.73 m/s
b)58.56°
Explanation:
given:
[tex]U_{y}[/tex]=20.7 m/s
t=49 s
a=0.3
solution:
a)
[tex]U_{y}[/tex] = [tex]V_{y}[/tex] = 20.7 m/s
[tex]U_{x}[/tex] =0
[tex]V_{x}[/tex] = [tex]U_{x}[/tex] + at
= 0 + 49 x 0.3
= 14.7 m/s
magnitude:
=[tex]\sqrt{V_{x} ^{2}+V_{y} ^{2} }[/tex]
=24.73 m/s
b)
angle =tan-1 [tex](\frac{V_{y} }{V_{x} } )[/tex]
= 58.56°
a)The magnitude of the satellite's velocity when the thruster turns off will be 24.73 m/sec.
b) The direction in terms of angle of the satellite's velocity when the thruster turns off will be 58.56°.
The change of displacement with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
In the y direction initial and the final velocity are equal;
[tex]\rm U_y= V_y = 20.7 \ m/sec[/tex]
From first equation of motion;
[tex]\rm V_x= U_X +at \\\\ 0 = V_x=49 \times 0.3 \\\\ V_x=14.7 \ m/sec[/tex]
The resultant velocity in the x and y direction will be;
[tex]\rm V = \sqrt{V_x^2+V_y^2} \\\\ V=24.73 \ m/sec[/tex]
The resultant angle is found as;
[tex]\rm \theta =tan^{-1} \frac{V_y}{V_x} \\\\\ \theta =58.56 ^0[/tex]
Hence,the magnitude of the satellite's velocity and the direction in terms of angle will be 24.73 m/sec and 58.56° respectively.
To learn more about the velocity refer to the link;
https://brainly.com/question/862972
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