Question

If velocity of the particle is given by v= √x
, where x
denotes the position of the particle and initially particle was
at x = 4, then which of the following are correct?

At t = 2 s, the position of the particle is x = 9

Particle's acceleration at t = 2 sec, is 1 m/s^2?

Particle's acceleration is 1/2m/s’^2 throughout the motion

Particle will never go in negative direction from its starting
position​

Answer 1

Answer:

Particle will never go in negative direction from its starting position.

Hence fourth option is correct

Explanation:

Given : velocity of particle v=  $\sqrt{x}$

            initially particle is at x=4

velocity  can be written as = $\frac{dx}{dt}$

   so ,

$\sqrt{x} = \frac{dx}{dt} \\\\dt = \frac{dx}{\sqrt{x} }$

integrating

$\int\limits^t_0 \, dt = \int\limits^x_0 {\frac{1}{\sqrt{x} } } \, dx \\\\t= \sqrt{x} +c$

initially at x= 4 ,t=0

 $0= \sqrt{4} + c\\\\c= -2$

so

$\\\\t=\sqrt{x} -2$

at t= 2

   $2 = \sqrt{x} -2\\\\\sqrt{x} = 4\\\\x= 16$

so first  statement is wrong

acceleration at t= 2

a= $\frac{dv}{dt}$

$a = \frac{d^{2} (t^{2} -4t+2)}{dt^{2} }\\ \\a= 2 m/s^{2}$

 acceleration is constant throughout the motion  a= 2m/s²

hence we can see  that starting first three option are not correct.

also particle cannot go to the negative direction until a back force is experienced which is not present in this case so last option is correct.