A spacecraft is traveling with a velocity of v0x = 5040 m/s along the +x direction. Two engines are turned on for a time of 801 s. One engine gives the spacecraft an acceleration in the +x direction of ax = 1.04 m/s2, while the other gives it an acceleration in the +y direction of ay = 8.70 m/s2. At the end of the firing, what is a) vx and b) vy?
To find the final velocities in the x and y directions, we can use the following equations:
a) vx = v0x + ax * t
b) vy = v0y + ay * t
Let's calculate each value step by step:
a) vx = v0x + ax * t
Given:
- v0x = 5040 m/s (initial velocity in the x direction)
- ax = 1.04 m/s^2 (acceleration in the x direction)
- t = 801 s (time interval)
Using the equation, we substitute the given values:
vx = 5040 m/s + 1.04 m/s^2 * 801 s
vx = 5040 m/s + 834.24 m/s
vx = 5874.24 m/s
Therefore, the final velocity in the x direction, vx, is 5874.24 m/s.
b) vy = v0y + ay * t
Given:
- ay = 8.70 m/s^2 (acceleration in the y direction)
- t = 801 s (time interval)
However, we don't have the initial velocity in the y direction, v0y. Since it is not provided in the given information, we assume that the spacecraft starts with zero initial velocity in the y direction (v0y = 0 m/s).
Using the equation, we substitute the given values:
vy = 0 m/s + 8.70 m/s^2 * 801 s
vy = 0 m/s + 6967.70 m/s
vy = 6967.70 m/s
Therefore, the final velocity in the y direction, vy, is 6967.70 m/s.