A) \[2s\]
B) \[1s\]
C) \[3/2s\]
D) \[1/2s\]
Correct Answer: D
Solution :
[d] From constant relations, we can see that acceleration of block B is \[{{a}_{B}}=\left( \frac{{{a}_{A}}+{{a}_{C}}}{2} \right)\] with proper signs Hence, \[{{a}_{B}}=\left( \frac{3-12t}{2} \right)=1.5-6t\] or \[\frac{d{{v}_{B}}}{dt}=1.5-6t\] or \[\int\limits_{0}^{{{v}_{B}}}{d{{v}_{B}}=\int\limits_{0}^{1}{(1.5-6t)dt}}\] or \[{{v}_{B}}=\frac{1}{5}t-3{{t}^{2}}\] or \[{{v}_{B}}=0\] at \[t=\frac{1}{2}s\]
You need to login to perform this action.
You will be redirected in
3 sec
