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

Question

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 = 4 m
(A) 6 sec
(B) 4 sec
(C)3 sec
(D) he can’t reach x = 4 m​

in progress 0
Genesis 5 months 2021-06-23T04:56:06+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-06-23T04:57:28+00:00

    Answer:

     \bf \huge \red{(A) \: 6sec}

    Explanation:

     \bf \: We \: know \: that,

      \bf \: v=d \times dt

     \bf \: so,

     \bf1x=d \times dt

     \bf \: Then,

     \bf∫dt=∫xdx

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

     \bf \color{red}=>x=6sec

    0
    2021-06-23T04:57:45+00:00

    \maltese\:\underline{\underline{\sf AnsWer :}}\:\maltese

    Given;

    \longrightarrow\:\:\tt v = \dfrac{1}{x} \\

    Also we know that velocity is written as;

    \longrightarrow\:\:\tt v = \dfrac{dx}{dt} \\

    Now,

    \longrightarrow\:\:\tt  \dfrac{dx}{dt} = \dfrac{1}{x} \\

    By cross multiplying we get :

    \longrightarrow\:\:\tt  dt = x.dx \\

    Now, integrate both the sides :

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

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

    \longrightarrow\:\:\displaystyle  \tt t= \dfrac{ {(4)}^{2} }{2}   -  \frac{ {(2)}^{2} }{2} \\

    \longrightarrow\:\:\displaystyle  \tt t= \dfrac{ 16 }{2}   -  \frac{ 4 }{2} \\

    \longrightarrow\:\:\displaystyle  \tt t= \dfrac{ 16 - 4 }{2}  \\

    \longrightarrow\:\:\displaystyle  \tt t= \dfrac{ 12 }{2}  \\

    \longrightarrow\:\: \underline{ \underline{\displaystyle  \tt t= 6 \: s }}\\

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

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