Asked by Anonymous
Two particles are moving in a plane; one has the velocity component vix=1m/s, viy=5m/s and the other has the velocity components v2x=4m/s and v2y=3m/s if both started from the same point what is the angle between their paths
Answers
Answered by
MathMate
calculate the sines and cosines of the angles ....
sintheta1=5/sqrt26
cosTheta1=1/sqrt26
sinTheta2=3/5
cosTheta2=4/5 check all those...
Now,
Sin(A-B)=sianAcosB - CosBsinA
then compute aine (A B), and finally, the arcsin(A-B) which is the angle
sintheta1=5/sqrt26
cosTheta1=1/sqrt26
sinTheta2=3/5
cosTheta2=4/5 check all those...
Now,
Sin(A-B)=sianAcosB - CosBsinA
then compute aine (A B), and finally, the arcsin(A-B) which is the angle
Answered by
Henry
V1 = 1 + 5i = 5.1m/s[78.7o].
V2 = 4 + 3i = 5m/s[36.9o].
78.7 - 36.9 = 41.8o between V1 and V2.
V2 = 4 + 3i = 5m/s[36.9o].
78.7 - 36.9 = 41.8o between V1 and V2.
Answered by
MathMate
Both solutions of the impersonator of MathMate and Henry give the same correct answer of 41.820°.
By the way, for your information, impersonation is a serious offence at Jiskha, liable to be banned.
By the way, for your information, impersonation is a serious offence at Jiskha, liable to be banned.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.