Asked by Harold
An asteroid heads for Earth at 12 km/s. In addition, a NASA space team is able to attach a rocket booster to the asteroid. The rocket booster fires for 10 min after which the asteroid is moving at 28 degrees to its original path to a speed of 20km/s. What is its average accleration for acceleartion at (x) and acceleartion at(y).
For acceleration (x) I got 12.9 m/s^2,but I am not sure about that or acceleartion at (y). Any help would be greatly appreciated.
For acceleration (x) I got 12.9 m/s^2,but I am not sure about that or acceleartion at (y). Any help would be greatly appreciated.
Answers
Answered by
bobpursley
acceleration= Vf-Vi /time
But draw the vector diagram, you see that Vf-Vi is the third vector in the vector triangle, and the law of cosines lets it be found...
(Vf-vi)^2=Vf^2+ Vi^2 - 2vf*Vi cos 28
acceleration then is (Vf-Vi) magnitude, divide by time in seconds.
But draw the vector diagram, you see that Vf-Vi is the third vector in the vector triangle, and the law of cosines lets it be found...
(Vf-vi)^2=Vf^2+ Vi^2 - 2vf*Vi cos 28
acceleration then is (Vf-Vi) magnitude, divide by time in seconds.
Answered by
Harold
I have no idea what you mean by this is my v initial 12km/s and my v final 20km/s
Answered by
bobpursley
Nope, they have direction associated with them. Vf-Vi is a vector, it is NOT 20-12
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