The position ' x ' of a particle varies with a time as x = at² - bt³ where ' a ' and ' b ' are constants. The acceleration of the particle will be zero at (a) 2a/3b (b) a/b (c) a/3b (d) zero

Answer: (c). v = dx/dt = 2at - 3bt², ap = dv/dt = 2a - 6bt. 2a - 6bt = 0 ⇒ t = a/3b So the answer is (c) a/3b.

The position ' x ' of a particle varies with a time as x = at² - bt³… | MHT CET 2023 (14 May, Shift 2)

Q15MHT CET 2023 · 14 MAY · SHIFT 2Easy

Velocity, Acceleration & Variable Motion

The position ' $x$ ' of a particle varies with a time as $x = at^2 - bt^3$ where ' $a$ ' and ' $b$ ' are constants. The acceleration of the particle will be zero at

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Answer: (c)

$v = \frac{dx}{dt} = 2at - 3bt^2$ , $a_{p} = \frac{dv}{dt} = 2a - 6bt$ .
$2a - 6bt = 0 \Rightarrow t = \frac{a}{3b}$
So the answer is (c) $\frac{a}{3b}$ .

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Velocity, Acceleration & Variable Motion question from MHT CET 2023 (14 May, Shift 2)

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