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:

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Kinematics

The relation between time '

t

' and distance '

x

' is

t = α x^{2} + β x

, where

α

and

β

are constants. The relation between acceleration (a) and velocity (v) is:

(A) $a = - 2 α v^{3}$ ✓ Correct
(B) $a = - 5 α v^{5}$
(C) $a = - 3 α v^{2}$
(D) $a = - 4 α v^{4}$

Answer: (A)

Solution

Given

t = α x^{2} + β x

. Differentiating with respect to

x

\frac{d t}{d x} = 2 α x + β ⟹ v = \frac{d x}{d t} = \frac{1}{2 α x + β}

Acceleration:

a = \frac{d v}{d t} = \frac{d v}{d x} \cdot v

Since

v = (2 α x + β)^{- 1}

\frac{d v}{d x} = - \frac{2 α}{( 2 α x + β ) ^{2}} = - 2 α v^{2}

a = v \cdot (- 2 α v^{2}) = - 2 α v^{3}

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option

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