Question

The displacement 'X (in meter) of a particle of mass
'm (in kg) moving in one dimension under the action
of a force, is related to time 't (in sec) by t = x +3.
The displacement of the particle when its velocity is
zero, will be​

Answer 1

Answer :

Option => B (Zero)

Explanation :

Correct Question :

The displacement x in metre of a particle of mass m Kg moving in one dimension under the action of a force is related to the time t in seconds by the equation: [t = √x + 3 ] . Then, the displacement of the particle, when its velocity is zero, is:

(a) 3 m.

(b) zero.

(c) 6 m.

(d) None of these.

Solution :

According to your question :

$\implies \tt t = \sqrt{x+3} = \sqrt{x} = t-3......(Eq1)$

Squaring on both sides of Equation 1 :

$\tt \implies x = (t-3)^2)\\ \\\implies t^2 +9 -6t.........Eq(2)$

Let the "v" be the velocity then,

$\tt \implies v = \dfrac{dx}{dt} = \dfrac{d}{dt}\quad (x)\\ \\\implies \dfrac{d}{dt}(t^2-6t+9)\\ \\\implies 2t-6....Eq(3)$

When,

$\tt \implies v = 0,2t-6=0\\ \\\implies t = 3sec$

Therefore, From Eqs 1, 2 and 3 :

$\implies \tt x = 3^2+9-3 \times 6\\ \\\implies 18-18\\ \\\implies 0\\ \\ \\\bigstar \textbf{\underline{\underline{Hence, The Answer is = Option B(0)}}}$