Question

The position (x) of a particle moving along x axis varies with time t as t=√3-3, where all quantities are in SI units. If displacement of the particle in time duration t=0 to t=5 s, is D, then find value of (D - 50) in m.


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Answer 1

Explanation:

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$\implies\rm t = \sqrt{x} -3$

•• differentiating both sides.

$\implies \rm dt = \dfrac{1}{2}\times \dfrac{1}{\sqrt{x}}.dx - 0$

$\implies \rm \dfrac{dx}{dt}= 2\sqrt{x}$

$\implies \bf v = 2\sqrt{x}$

We know that, a = v.dv/dx

$\implies \rm v = 2\sqrt{x}$

$\implies \rm dv= 2\times \dfrac{1}{2\: \sqrt{x}} .dx$

$\implies \rm \dfrac{dv}{dx}= \dfrac{1}{\sqrt{x}}$

$\to\rm v.\dfrac{dv}{dx} = 2\sqrt{x}\times \dfrac{1}{\sqrt{x}}$

$\to \bf a = 2 ms^{-2}$

★ As acceleration is constant, displacement by equation of motion:

s = ut +½(a)t²

D = 0 + ½(2){(5)²-(0)²}

D = 25

》 D - 50 = 25 - 50 = -25m

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Answer 2

Given: The position (x) of a particle moving along x axis varies with time t as t = √x - 3 and the displacement of the particle in time duration t=0 to t=5 s is D.

To find: The value of (D - 50) in m.

Solution:

• Since the time varies as t = √x - 3, we calculate the positions for the time 0s and for 5s.
• Thus, for t=0s, we have,

        $0 = \sqrt{x} - 3$

        $x = \sqrt{3} m$

• For t=5s, we have,

        $5 = \sqrt{x} - 3$

        $x = \sqrt{2}m$

• If the displacement is D, then D is the difference between √2m and √3m.

        $\sqrt{2} - \sqrt{3} = - 0.318m$

• We find the difference between the displacements to find D, that lies between the given time interval of 0s to 5s.
• So,

       $D - 50 = -0.318 - 50$

                   $= -50.318m$

Therefore, the value of D is -50.318m.