Question

In the figure, acceleration of A is ${\overrightarrow{a}}_A=15\hat{i}+15\hat{j}$, then acceleration of B is (A remains in contact with B)

Answer 1

According to the figure

$a_{Ax}$ is actually $a_{Ay}$and vice-versa

Now,

$a_{Ax}=15$

$a_{Ay}=15$

$a_{Ax}={\left(a_{By}\right)}_{\perp }$

$a_{Ax}\cos {53}^0-a_{Ay}\cos {37}^0$

$=a_B\cos {53}^0$

$a_B=-5m/s$

$a_B=5\hat{i}$

Hence, this is the required solution