A man moves along the x-axis such that its velocity is v =1/x. If he is initially at x = 2 m, find the time when
he reaches x

2 thoughts on “A man moves along the x-axis such that its velocity is v =1/x. If he is initially at x = 2 m, find the time when<br />he reaches x”

1. Answer:

   [tex] \bf \huge \red{(A) \: 6sec}[/tex]

   Explanation:

   [tex] \bf \: We \: know \: that,[/tex]

   [tex] \bf \: v=d \times dt[/tex]

   [tex] \bf \: so,[/tex]

   [tex] \bf1x=d \times dt[/tex]

   [tex] \bf \: Then,[/tex]

   [tex] \bf∫dt=∫xdx[/tex]

   [tex]\bf\displaystyle \ t=\left. \dfrac{ {x}^{2} }{2} \right| _{2}^{4} \\ [/tex]

   [tex] \bf \color{red}=>x=6sec[/tex]

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2. [tex]\maltese\:\underline{\underline{\sf AnsWer :}}\:\maltese[/tex]

   Given;

   [tex]\longrightarrow\:\:\tt v = \dfrac{1}{x} \\ [/tex]

   Also we know that velocity is written as;

   [tex]\longrightarrow\:\:\tt v = \dfrac{dx}{dt} \\[/tex]

   Now,

   [tex]\longrightarrow\:\:\tt \dfrac{dx}{dt} = \dfrac{1}{x} \\[/tex]

   By cross multiplying we get :

   [tex]\longrightarrow\:\:\tt dt = x.dx \\[/tex]

   Now, integrate both the sides :

   [tex]\longrightarrow\:\:\displaystyle \tt\int\limits_{0}^{t}dt=\int\limits_{2}^{4}
   x.dx \\ [/tex]

   [tex]\longrightarrow\:\:\displaystyle \tt t=\left. \dfrac{ {x}^{2} }{2} \right| _{2}^{4} \\ [/tex]

   [tex]\longrightarrow\:\:\displaystyle \tt t= \dfrac{ {(4)}^{2} }{2} – \frac{ {(2)}^{2} }{2} \\ [/tex]

   [tex]\longrightarrow\:\:\displaystyle \tt t= \dfrac{ 16 }{2} – \frac{ 4 }{2} \\ [/tex]

   [tex]\longrightarrow\:\:\displaystyle \tt t= \dfrac{ 16 – 4 }{2} \\ [/tex]

   [tex]\longrightarrow\:\:\displaystyle \tt t= \dfrac{ 12 }{2} \\ [/tex]

   [tex]\longrightarrow\:\: \underline{ \underline{\displaystyle \tt t= 6 \: s }}\\ [/tex]

   Hence, the option (A) t = 6 seconds is a correct answer.

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