Question

On a spacecraft two engines fire for a time of 399 s. One gives the craft an acceleration in the x direction of an x = 4.63 m/s2, while the other produces an acceleration in the y direction of ay = 8.15 m/s2. At the end of the firing period, the craft has velocity components of vx = 3160 m/s and vy = 4020 m/s. Find (a) the magnitude and (b) the direction of the initial velocity. Express the direction as an angle with respect to the +x axis.

Answer 1

Answer:

Explanation:

We can determine the initial velocity of the spacecraft using the data provided using the equations:

$v_0x = v_x - a_x*t$

$v_0y = v_y - a_y*t$

Replacing with the known data, we have that:

$v_0x = v_x - a_x* t= 3160 m/s - 4,63 m/s^2 * (399 s) = 1312.63 m/s$

$v_0y = v_y - a_y * t = 4020 m/s - 8,15 m/s^2 *( 399 s ) =768,15 m/s$

Once we know the components vx and vy of the initial velocity, we can apply trigonometry to determine the magnitude and direction of the vector initial velocity, as follows. First, we use the Pythagorean theorem:

$V_0 =\sqrt{(v_0x^2)+((v_0y^2)}= \sqrt{(1312,63 m/s)^2+(768,15 m/s)^2}=1520,87 m/s$

Then, we find the angle with respect to the +x axis using the definition of tangent:

$Ф=tan^-1(v_0y/v_0x) =tan^-1((768,15m/s)/(1312,63 m/s))= 30,11°$

I hope everything was clear with my explanation. If I can help you any futher, please let me know. Have a great day :D