Question

A spaceship is traveling at a constant velocity of →v(t) = 250.0ˆi m/s when its rockets fire, giving it an acceleration of →a(t) = (3.0ˆi + 4.0ˆk) m/s². What is its velocity 5 s after the rockets fire?

a) 265.0ˆi m/s
b) 275.0ˆi m/s
c) 285.0ˆi m/s
d) 295.0ˆi m/s

Answer 1

The velocity of the spaceship 5 seconds after the rockets fire is 265.0î m/s.

Step-by-step explanation:

To find the velocity of the spaceship 5 seconds after the rockets fire, we need to integrate the acceleration function over time. Given that the initial velocity is 250.0î m/s and the acceleration is (3.0î + 4.0k) m/s², we can integrate the x-component of the acceleration to find the change in velocity in the x-direction. By integrating 3.0î with respect to time, we get 3.0tî. Plugging in t = 5, we get 3.0(5)î = 15î m/s. Adding this to the initial velocity gives us 250.0î m/s + 15î m/s = 265.0î m/s.