In the figure shown the acceleration of A is, `vec(a)_(A) = 15 hat (i) + 15 hat (j)` then the acceleration of B is (A remains in contact in B)

In fig, the acceleration of A is vec(a)_(A)=15hat(i)+15hat(j) . Then the acceleration of B is (A remains in constact with B)

In fig, the acceleration of A is vec(a)_(A)=15hat(i)+15hat(j) . Then the acceleration of B is (A remains in constact with B)

In the figure shown the accelerationof A is a_(A) = (15 hati + 15 hatj) . Then the acceleration of B is (A remain in contact with B)

In the figure shown the accelerationof A is a_(A) = (15 hati + 15 hatj) . Then the acceleration of B is (A remain in contact with B)

In the figure shown the acceleration of A is, veca =15hati+15hatj then the acceleration of B is: (A remains in contact with B)

The velocities of two particles A and B of same mass are vec V_A = a hat i and vec V_B = b hat j where a and b are constants. The acceleration of particle A is (2 a hat i+ 4 b hat j) and acceleration of particle B is (a hat i- 4 b hat j) (in m//s^2) . The centre of mass of two particle will move in :

In the figure shown the velocity of A with respect to ground is (20hat(i) + 15hat(j)) m//s , then what will be the magnitude of velocity of the wedge 'B' with respect to ground frame. (A remains in contact with B)

In the figure shown the velocity of A with respect to ground is (20hat(i) + 15hat(j)) m//s , then what will be the magnitude of velocity of the wedge 'B' with respect to ground frame. (A remains in contact with B)

Find the scalar projection of : vec(a)=hat(i)-hat(j) on vec(b)=hat(i)+hat(j)

If vec(A) = 3 hat(i) - 4 hat(j) and vec(B) = 2 hat(i) + 16 hat(j) then the magnitude and direction of vec(A) + vec(B) will be