Question

The speed v of a particle moving along the x-axis, is
given by v2 = 10x - 5x2. The magnitude of its
acceleration when it is at x= 1 m is​

Answer 1

Explanation:

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Answer 2

Answer:

The correct answer is equal to $0m/sec^{2}$.

Explanation:

$v^{2}=10x-5x^{2}$

Differentiate both sides with respect to time,

$\frac{d}{dt} (v^{2})=\frac{d}{dt} (10x-5x^{2})$

$2v\frac{dv}{dt}=10\frac{dx}{dt}-10x\frac{dx}{dt}$

$2va=10v-10xv$                                ∴($a=\frac{dv}{dt}$ and $v=\frac{dx}{dt}$)

a = 5 - 5x

The acceleration of a particle when it is at x = 1m moving along the x-axis,

a = 5 - 5 × 1

a = 0$m/sec^{2}$

So, the acceleration of a particle when it is at x = 1m moving along the x-axis is equal to $0m/sec^{2}$.