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)

Options

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

Answer

D

Explanation

Let `a_(B) = a hati`
Then, `a_(AB) = a_(A) - a_(B)`
`= (15 - a) hati + 15 hatj`

Since `a_(AB)` is along the plane as shown in figure.
`:. tan 37 ^(@) = (3)/(4) = (15)/(15 - a)`
Solving this equation , we get `a = - 5`
or `a_(B) = (- 5 hati)`