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A particle has two equal acclerations in two given direction.If one of the acceleration is half then the angle which the resultan makes with the other is also half.The angle between the acceleration is???
Dear Banjit, Let,'a' be the magnitude of acceleration, 'x' be the angle between accelerations, and 'y' be the angle between resultant and one of the acceleration. Then, tan y = a sin x / (a + acos x) [ Standard formula in vectors] tany = sinx/ (1+ cos x)........................(1) If 'a' becomes 'a/2' then 'y' becomes 'y/2' So, tan(y/2) = (a/2) sin x / (a + (a/2) cosx) tan (y/2) = sinx / (2+cos x) Also, tany = 2tan(y/2) /(1 - tan2 (y/2) ) Putting value of tan(y/2) in above eq. will give- tan y = 2sinx (2+ cos x) / (2cos2x + 4cos x + 3)...................(2) On comparing eq. (1) and (2), we get- 2(1 + cosx) (2+ cos x) = (2cos2x + 4cos x + 3) or, cos x = 1/2 or, x= 600 [ANS] Please feel free to post as many doubts on our discussion forum as you can. If you find any questionDifficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best Banjit!!!Regards,Askiitians ExpertsPriyansh Bajaj
Dear Banjit,
Let,'a' be the magnitude of acceleration, 'x' be the angle between accelerations, and 'y' be the angle between resultant and one of the acceleration. Then, tan y = a sin x / (a + acos x) [ Standard formula in vectors] tany = sinx/ (1+ cos x)........................(1) If 'a' becomes 'a/2' then 'y' becomes 'y/2' So, tan(y/2) = (a/2) sin x / (a + (a/2) cosx) tan (y/2) = sinx / (2+cos x) Also, tany = 2tan(y/2) /(1 - tan2 (y/2) ) Putting value of tan(y/2) in above eq. will give- tan y = 2sinx (2+ cos x) / (2cos2x + 4cos x + 3)...................(2) On comparing eq. (1) and (2), we get- 2(1 + cosx) (2+ cos x) = (2cos2x + 4cos x + 3) or, cos x = 1/2 or, x= 600 [ANS] Please feel free to post as many doubts on our discussion forum as you can. If you find any questionDifficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best Banjit!!!Regards,Askiitians ExpertsPriyansh Bajaj
Let,'a' be the magnitude of acceleration, 'x' be the angle between accelerations, and 'y' be the angle between resultant and one of the acceleration.
Then, tan y = a sin x / (a + acos x) [ Standard formula in vectors]
tany = sinx/ (1+ cos x)........................(1)
If 'a' becomes 'a/2' then 'y' becomes 'y/2'
So, tan(y/2) = (a/2) sin x / (a + (a/2) cosx)
tan (y/2) = sinx / (2+cos x)
Also, tany = 2tan(y/2) /(1 - tan2 (y/2) )
Putting value of tan(y/2) in above eq. will give-
tan y = 2sinx (2+ cos x) / (2cos2x + 4cos x + 3)...................(2)
On comparing eq. (1) and (2), we get-
2(1 + cosx) (2+ cos x) = (2cos2x + 4cos x + 3)
or, cos x = 1/2
or, x= 600 [ANS]
Please feel free to post as many doubts on our discussion forum as you can. If you find any questionDifficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best Banjit!!!Regards,Askiitians ExpertsPriyansh Bajaj
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