The relation between time t and distance x is t = ax^2 + bx where a and b are constants. The acceleration is: - Physics MCQ AIEEE 2026 | ExamTest.live

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The relation between time t and distance x is t = a x^{2} + b x where a and b are constants. The acceleration is:

Solution & Step-by-step Explanation

Given t = a x^{2} + b x .Differentiating w.r.t. x : \frac{d t}{d x} = 2 a x + b Since v = \frac{d x}{d t} : v = \frac{1}{2 a x + b} = (2 a x + b)^{- 1} Differentiating velocity w.r.t. t to find acceleration (f): f = \frac{d v}{d t} = \frac{d v}{d x} \cdot \frac{d x}{d t} \frac{d v}{d x} = - 1 (2 a x + b)^{- 2} \cdot (2 a) = - 2 a (2 a x + b)^{- 2} Since v = (2 a x + b)^{- 1}, then (2 a x + b)^{- 2} = v^{2} . \frac{d v}{d x} = - 2 a v^{2} Acceleration f = (- 2 a v^{2}) \cdot v = - 2 a v^{3} .

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The relation between time t and distance x is t = a x^{2} + b x where a and b are constants. The acceleration is:

- 2 ab v^{2}

2 b v^{3}

- 2 a v^{3}

2 a v^{2}

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