On a spacecraft two engines fire for a time of 581 s. One gives the craft an acceleration in the x direction of ax = 5.21 m/s2, while the other produces an acceleration in the y direction of ay = 7.34 m/s2. At the end of the firing period, the craft has velocity components of vx = 3753 m/s and vy = 4818 m/s. Calculate the magnitude of the initial velocity.
To solve this problem, we need to break it down into two separate problems: one in the x-direction and one in the y-direction.
In the x-direction, we can use the first equation of motion:
vf = vi + at
Given that the acceleration (ax) is 5.21 m/s^2, the time (t) is 581 s, and the final velocity (vf) in the x-direction (vx) is 3753 m/s, we can rearrange the equation to solve for the initial velocity (vi):
vi = vf - at
Plugging in the values, we get:
vi = 3753 m/s - (5.21 m/s^2)(581 s)
vi ≈ 3753 m/s - 3026.01 m/s
vi ≈ 727 m/s
So, the initial velocity in the x-direction (vxi) is approximately 727 m/s.
Similarly, in the y-direction, we can use the same equation of motion:
vf = vi + at
Given that the acceleration (ay) is 7.34 m/s^2, the time (t) is 581 s, and the final velocity (vf) in the y-direction (vy) is 4818 m/s, we can rearrange the equation to solve for the initial velocity (vi):
vi = vf - at
Plugging in the values, we get:
vi = 4818 m/s - (7.34 m/s^2)(581 s)
vi ≈ 4818 m/s - 4260.54 m/s
vi ≈ 558 m/s
So, the initial velocity in the y-direction (viy) is approximately 558 m/s.
To find the magnitude of the initial velocity, we can use the Pythagorean theorem:
v = sqrt(vx^2 + vy^2)
Plugging in the values, we get:
v = sqrt((727 m/s)^2 + (558 m/s)^2)
v ≈ sqrt(529329 m^2/s^2 + 311364 m^2/s^2)
v ≈ sqrt(840693 m^2/s^2)
v ≈ 917 m/s
Therefore, the magnitude of the initial velocity is approximately 917 m/s.