A) \[\frac{{{v}^{2}}}{a}\]
B) \[\frac{{{v}^{2}}}{2a}\]
C) \[\frac{2{{v}^{2}}}{a}\]
D) \[\frac{{{v}^{2}}}{4a}\]
Correct Answer: B
Solution :
[b] Let x be the distance between the particles after t seconds. Then \[x=vt-\frac{1}{2}a{{t}^{2}}\] ...(1) For x to be maximum, \[\frac{dx}{dt}=0\] or \[v-at=0\] or \[t=\frac{v}{a}\] Substituting this value in (1), we get \[x=v\left( \frac{v}{a} \right)-\frac{1}{2}a{{\left( \frac{v}{a} \right)}^{2}}\] or \[x=\frac{{{v}^{2}}}{2a}\]You need to login to perform this action.
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