A particle is moving along x -axis with its position (x) varying with time (t) as x = α t⁴ + β t² + γ t + δ. The ratio of its initial velocity to its initial acceleration, respectively, is: (a) 2α: δ (b) γ: 2δ (c) 4α: β (d) γ: 2β

Answer: (d). v = 4α t³ + 2β t + γ, so at t = 0: v₀ = γ. a = 12α t² + 2β, so at t = 0: a₀ = 2β. So the answer is (d) γ: 2β.

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Q7NEET 2024Easy

Velocity, Acceleration & Variable Motion

A particle is moving along $x$ -axis with its position (x) varying with time $(t)$ as $x = \alpha t^4 + \beta t^2 + \gamma t + \delta$ . The ratio of its initial velocity to its initial acceleration, respectively, is:

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Answer: (d)

$v = 4\alpha t^3 + 2\beta t + \gamma$ , so at $t = 0$ : $v_0 = \gamma$ .
$a = 12\alpha t^2 + 2\beta$ , so at $t = 0$ : $a_0 = 2\beta$ .
So the answer is (d) $\gamma : 2\beta$ .

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Velocity, Acceleration & Variable Motion question from NEET 2024

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