A) \[{{v}^{2}}<ax\]
B) \[{{v}^{2}}<2ax\]
C) \[{{v}^{2}}\ge 2ax\]
D) \[{{v}^{2}}=ax\]
Correct Answer: C
Solution :
Let the police catch the thief in time t Distance travelled in time t = vt According to the question. \[x+\frac{1}{2}a\,{{t}^{2}}=vt\] \[\frac{{{t}^{2}}}{2}-vt+x=0\] \[t=\frac{v\pm \sqrt{{{v}^{2}}-2ax}}{a}\] For real t, \[{{v}^{2}}-2ax\ge 0\] \[{{v}^{2}}\ge 2ax.\]You need to login to perform this action.
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