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​

Answer 1

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

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 :

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

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

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

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

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

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

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

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

Answer 2

Answer:

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

Explanation:

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

$\bf \: v=d \times dt$

$\bf \: so,$

$\bf1x=d \times dt$

$\bf \: Then,$

$\bf∫dt=∫xdx$

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

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