High School

A space shuttle is flying north at 6510 m/s. It then begins accelerating at [tex]28.4 \, \text{m/s}^2[/tex] directly northeast to properly align its path towards orbit around the Earth. After 60.0 seconds, what is the magnitude of its velocity?

Answer :

Final answer:

The magnitude of its velocity after accelerating northeast for 60 seconds is approximately 6724 m/s.

Explanation:

The initial velocity of the shuttle is 6510 m/s. Assuming the path of the shuttle as it accelerates towards northeast forms a right angled triangle with the initial path, you can calculate the final velocity using Pythagoras' theorem. The horizontal component is still 6510 m/s, but there is now a vertical component due to the acceleration. The vertical component is the acceleration (28.4 m/s2) times time (60.0s), which equals 1704 m/s. Now, use Pythagoras' theorem to find the resultant velocity (i.e., the magnitude of its velocity) that is given by the square root of (65102 + 17042) which approximates to 6724 m/s.

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