A particle starts moving with initial velocity along the -axis from origin.

Question

A particle starts moving with initial velocity $4 {ms}^{- 1}$ along the $x$ -axis from origin. Its acceleration is varying with its position $x$ in parabolic nature as shown in figure. At $x = \sqrt{3}$ m, a tangent to the curve makes an angle $60 °$ with positive $x$ -axis as shown. Then at $x = \sqrt{3}$ m

Statement-I: Velocity $(v) = \sqrt{\sqrt{3} + 16}$ ${ms}^{- 1}$

Statement-II: Acceleration $(a) = 3 {ms}^{- 2}$

Statement-I: Velocity $(v) = \sqrt{\sqrt{3} + 16}$ ${ms}^{- 1}$

Statement-II: Acceleration $(a) = 3 {ms}^{- 2}$

Smart Achievers · Accepted Answer

(2)

$For a parabolic a - x graph, let$

$a = k x^{2}$

$60 °$

$\frac{d a}{d x} = \tan 60 ° = \sqrt{3}$

$2 k x = \sqrt{3}$

$2 k \sqrt{3} = \sqrt{3} \Rightarrow k = \frac{1}{2}$

$Therefore,$

$a = \frac{x^{2}}{2} = \frac{3}{2} m / s^{2}$

$So, Statement-II is false.$

$Now,$

$a = v \frac{d v}{d x}$

$v d v = \frac{x^{2}}{2} d x$

$Integrating from x = 0, v = 4 to x = \sqrt{3}, v = v,$

$\frac{v^{2} - 16}{2} = \int_{0}^{\sqrt{3}} \frac{x^{2}}{2} d x = \frac{\sqrt{3}}{2}$

$v^{2} = 16 + \sqrt{3}$

$v = \sqrt{16 + \sqrt{3}} m / s$

$Hence, Statement-I is true and Statement-II is false.$

Solution

1 Both Statement – I and Statement – II are true

2 Statement – I is true and Statement – II is false

3 Statement – I is false and Statement – II is true

4 Both Statement – I and Statement – II are false

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