The coordinates of a moving particle at any time t are given by x = α t ^ 3 and and y = β t ^ 3 . The speed of the particle at time t & is given by

Correct answer: D. 3 t ^ 2 α ^ 2 + β ^ 2 ∵ x = α t ^ 3 ∴ (d x)/(d t ) = 3 ^ 2 ⇒ v _ x = 3 ^ 2 Again y = β t ^ 3 ∴ (d y )/(d t ) ⇒ v _ y = 3 ^ 2 Since, v^ 2 = v _x^ 2 + v _y^ 2 v^ 2 = ( 3 ^ 2 ) ^ 2 + ( 3 ^ 2 ) ^ 2 = ( 3t ^ 2 ) ^ 2 ( α ^ 2 + β ^ 2 ) Or v = 3t ^ 2 α ^ 2 + β ^ 2

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Physics
Chapter: Motion in Two Dimensions

NTA Abhyas 2020 · Physics · Motion in Two Dimensions

The coordinates of a moving particle at any time t are given by x =...

The coordinates of a moving particle at any time

t

are given by

x = \alpha {t }^{3 \text{ and } }

and

y = \beta {t }^{3}

. The speed of the particle at time

t

& is given by

Options

A $3 t \sqrt{{\alpha}^{2} + {\beta}^{2}}$
B ${t }^{2} \sqrt{{\alpha}^{2} + {\beta}^{2}}$
C $\sqrt{{\alpha}^{2} + {\beta}^{2}}$
D $3 {t }^{2} \sqrt{{α}^{2} + {β}^{2}}$

Correct answer

D. $3 {t }^{2} \sqrt{{α}^{2} + {β}^{2}}$

Step-by-step solution

∵ x = \alpha {t }^{3}

∴ \dfrac{d x}{d t } = 3\alphat{}^{2} ⇒ {v}_{x} = 3\alphat{}^{2}

Again

y = \beta {t }^{3}

∴ \dfrac{d y }{d t } ⇒ {v}_{y } = 3\betat{}^{2}

Since,

v^{2} = {v}_x^{2} + {v}_y^{2}

v^{2} = ( 3\alphat{}^{2}{)}^{2} + ( 3\betat{}^{2}{)}^{2} = ( 3t{}^{2}{)}^{2} ( {\alpha}^{2} + {\beta}^{2} )

v = 3t{}^{2}\sqrt{{\alpha}^{2} + {\beta}^{2}}

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