Question

A spacecraft is traveling with a velocity of v0x = 6580 m/s along the +x direction. Two engines are turned on for a time of 907 s. One engine gives the spacecraft an acceleration in the +x direction of ax = 1.81 m/s2, while the other gives it an acceleration in the +y direction of ay = 8.05 m/s2. At the end of the firing, what is a) vx and b) vy?

Answer 1

`v_(x)=8222m/s\\v_(y)=7301m/s`

Step-by-step explanation:

From the exercise we know that

`v_(ox)=6580m/s\\a_(x)=1.81m/s^2\\a_(y)=8.05m/s^2`

`t=907s`

Knowing the following formula we can calculate the final velocity

`v=v_(o)+at`

`v_(x)=v_(ox)+a_(x)t`

`v_(x)=6580m/s+(1.81m/s^2)(907s)=8222m/s`

`v_(y)=v_(oy)+a_(y)t`

`v_(y)=(8.05m/s^2)(907s)=7301m/s`