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In figure, the acceleration of A is ${\vec{a}}_{A} = 15 \hat{i} + 15 \hat{j}$ . Then the acceleration of B is (A remains in contact with B)

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6i^

-15i^

-10i^

-5i^

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Correct option is D

From constraint, the acceleration of both block and wedge should be same in a direction perpendicular to the inclined plane as shown in figure.aA⊥=aB⊥,aAX=15,aAY=15aAXcos⁡53∘−aAYcos⁡37∘=aBcos⁡53∘or aB=−5ms−1 or a→B=−5i^

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In figure, the acceleration of A is a→A=15i^+15j^. Then the acceleration of B is (A remains in contact with B)

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In figure, the acceleration of A is ${\vec{a}}_{A} = 15 \hat{i} + 15 \hat{j}$ . Then the acceleration of B is (A remains in contact with B)