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Q.At what angle the two forces A + B and A - B act, so that their resultant is sqrt(3A^2 + B^2).

(OR)
A Car travelling at 9 ms^-1 accelerates and attains a speed of 27 ms^-1 in 5 seconds. Calculate the acceleration and distance covered in 5 seconds.
Himachal HpboseHPBOSE Himachal Pradesh Class 11 Board Exam 2021Subjective· 2mImportance★★★★★
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Using R^2 = P^2 + Q^2 + 2PQ cos(theta) with P = A+B and Q = A-B, and setting R^2 = 3A^2+B^2 as given, solving for cos(theta) gives cos(theta) = 1/2, so theta = 60 degrees.

Let the two vectors be P = (A + B) and Q = (A - B), with an angle theta between them (A and B are the magnitudes of two vectors being added/subtracted). Their resultant magnitude R is given by the parallelogram law of vector addition:

R^2 = P^2 + Q^2 + 2PQ cos(theta)

We are given R = sqrt(3A^2 + B^2), so R^2 = 3A^2 + B^2.

Now:

P^2 + Q^2 = (A+B)^2 + (A-B)^2 = (A^2 + 2AB + B^2) + (A^2 - 2AB + B^2) = 2A^2 + 2B^2

2PQ = 2(A+B)(A-B) = 2(A^2 - B^2)

Substituting into the resultant equation:

3A^2 + B^2 = (2A^2 + 2B^2) + 2(A^2 - B^2) cos(theta)

3A^2 + B^2 - 2A^2 - 2B^2 = 2(A^2 - B^2) cos(theta)

A^2 - B^2 = 2(A^2 - B^2) cos(theta)

Dividing both sides by (A^2 - B^2), assuming A is not equal to B:

cos(theta) = 1/2

theta = 60 degrees

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