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In the figure shown the acceleration of ...

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)

A

`6 hat(i)`

B

`- 15 hat (i)`

C

`- 10 hat (i)`

D

`- 5 hat (i)`

Text Solution

Verified by Experts

The correct Answer is:
D

From wedge constant
`(a_(A))_(bot)=(a_(B))_(bot)`
`a_(AX) cos 53^(@) -a_(AY) cos 37^(@) = a_(B) cos 53^(@)`
`a_(B) = -5 m//s rArr vec(a)_(B) = -5 hat(i)`
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Knowledge Check

  • 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)

    A
    `6hat(i)`
    B
    `-15hat(i)`
    C
    `-10hat(i)`
    D
    `-5hat(i)`

  • 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)

    A
    `5 hati`
    B
    `-15 hati`
    C
    `-10 hati`
    D
    `-5 hati`

  • 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 :

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