The relation between time ' t ' and distance ' x ' is t = α x² + β x, where α and β are constants. The relation between acceleration (a) and velocity (v) is: (a) a = -5α v⁵ (b) a = -3α v² (c) a = -2α v³ (d) a = -4α v⁴

Answer: (c). dt/dx = 2α x + β = 1/v Differentiate w.r.t. t: 2α v = -1/v²dv/dt a = dv/dt = -2α v³ So the answer is (c).

The relation between time ' t ' and distance ' x ' is t = α x² + β x… | JEE Main 2024 (31 Jan, Shift 1)

Q11JEE Main 2024 · 31 JAN · SHIFT 1Medium

Velocity, Acceleration & Variable Motion

The relation between time ' $t$ ' and distance ' $x$ ' is $t = \alpha x^2 + \beta x$ , where $\alpha$ and $\beta$ are constants. The relation between acceleration $(a)$ and velocity $(v)$ is:

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Answer: (c)

$\dfrac{dt}{dx} = 2\alpha x + \beta = \dfrac{1}{v}$
Differentiate w.r.t. $t$ : $2\alpha v = -\dfrac{1}{v^2}\dfrac{dv}{dt}$
$a = \dfrac{dv}{dt} = -2\alpha v^3$
So the answer is (c).

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