On a spacecraft two engines fire for a time of 560 s. One gives the craft an acceleration in the x direction of ax = 5.10 m/s2, while the other produces an acceleration in the y direction of ay = 7.30 m/s2. At the end of the firing period, the craft has velocity components of vx = 3700 m/s and vy = 5006 m/s. Find the magnitude and direction of the initial velocity. Express the direction as an angle with respect to the +x axis.
what is the magnitude in m/s?
what is the direction in ° (counterclockwise from the +x direction)
werhj
To find the magnitude and direction of the initial velocity, we can use the equations of motion for the spacecraft.
First, let's find the initial velocity components. We know that the craft starts from rest, so the initial velocity in both the x and y directions would be zero.
Next, we can find the final velocity components in the x and y directions using the given information. The final velocity components are vx = 3700 m/s and vy = 5006 m/s.
Now, we can find the magnitude of the final velocity using the Pythagorean theorem:
Magnitude of the final velocity (v) = √(vx^2 + vy^2)
= √((3700 m/s)^2 + (5006 m/s)^2)
= √(13690000 m^2/s^2 + 25060036 m^2/s^2)
= √(38750036 m^2/s^2)
≈ 6227.6 m/s
So, the magnitude of the initial velocity is approximately 6227.6 m/s.
To find the direction of the initial velocity, we can use trigonometry.
Direction (θ) = atan(vy / vx)
= atan(5006 m/s / 3700 m/s)
≈ atan(1.3551)
≈ 52.25°
Therefore, the direction of the initial velocity is approximately 52.25° counterclockwise from the +x direction.