A) 4
B) 2
C) 1
D) 0
Correct Answer: D
Solution :
\[t=\sqrt{x}+2\] \[\sqrt{x}=t-2\] \[\operatorname{x}= {{\left( t - 2 \right)}^{2}}\] (Squaring both sides) \[\operatorname{x}= {{t}^{2}}+4-4t\] ? (i) Now velocity, \[v=\frac{dx}{dt}=\frac{d}{dt}({{t}^{2}}+4-4t)=2\,t-4\] \[v=0\,\,\Rightarrow \,\,2t\,-4=0\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,t=2\,\,s\] Putting \[r = 2\,\,s\] in ... (ii) \[\operatorname{x}=4+ 4-8=0\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\operatorname{Displacement} = 0\]You need to login to perform this action.
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