NEET Physics Question - Kinematics - Calculus Based Questions

If the velocity of a particle is given by \[v=(180-16x)^{1/2} ms^{-1}\], then its acceleration will be

zero

\[16 ms^{-2}\]

\[-8 ms^{-2}\]

\[4 ms^{-2}\]

Given the velocity:

$v = \sqrt{180 - 16 x} m/s$

To find the acceleration, use the chain rule:

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

Differentiate $v$ with respect to $x$ :

$\frac{dv}{dx} = \frac{-8}{\sqrt{180 - 16x}}$

Now, substitute $v = \sqrt{180 - 16x}$

$a = - 8 {m/s}^{2}$

Thus, the acceleration is $a = -8 \, \text{m/s}^2$

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