A) zero
B) \[10\sqrt{2}m/s\]
C) \[10\sqrt{3}m/s\]
D) \[10\,m/s\]
Correct Answer: D
Solution :
Let the particle be moving in counter clockwise direction. The initial velocity of the particle \[{{\vec{v}}_{1}}=v\hat{j}\] The velocity of the particle after time t is \[v(\cos {{60}^{0}}\hat{j}-\sin {{60}^{o}}\hat{i})=\frac{v}{2}(j-\sqrt{3}\hat{i})\] So, the change in velocity \[|\Delta v|=|{{\vec{v}}_{2}}-{{\vec{v}}_{1}}|=v\left| \frac{1}{2}(\hat{j}-\sqrt{3}\hat{i})-\hat{j} \right|\] \[=\frac{10}{2}\left| -\hat{j}-\sqrt{3}\hat{i} \right|=5\sqrt{{{(-1)}^{2}}+{{(-\sqrt{3})}^{2}}}=10m/s\]You need to login to perform this action.
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