Question

A particle is moving on the x-axis such that its velocity varies with the x coordinate according to the relation v = 2x + 1. Find out the acceleration of the particle as a function of x.​

Answer 1

Answer :-

4x + 2

Explanation :-

Given conditions,

v = 2x + 1

Differentiating both sides with respect to time,

dv/dt = 2(dx/dt) + 0

Since, dv/dt = a and dx/dt = v, we will get,

a = 2v

= 2(2x + 1)

= 4x + 2

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Answer 2

The acceleration of the particle as a function of x is 4x + 2

Given :

A particle is moving on the x-axis such that its velocity varies with the x coordinate according to the relation v = 2x + 1

To find :

The acceleration of the particle as a function of x

Solution :

Step 1 of 2 :

Write down the given relation

Here it is given that the particle is moving on the x-axis such that its velocity varies with the x coordinate according to the relation v = 2x + 1

Step 2 of 2 :

Calculate acceleration of the particle as a function of x

$\displaystyle \sf v = 2x + 1$

Differentiating both sides with respect to x we get

$\displaystyle \sf \frac{dv}{dx} = \frac{d}{dx} (2x + 1)$

$\displaystyle \sf{ \implies }\frac{dv}{dx} = \frac{d}{dx} (2x) + \frac{d}{dx} ( 1)$

$\displaystyle \sf{ \implies }\frac{dv}{dx} = 2\frac{d}{dx} (x) + 0$

$\displaystyle \sf{ \implies }\frac{dv}{dx} = 2$

Hence the acceleration of the particle

$\displaystyle \sf{ = }v\frac{dv}{dx}$

$\displaystyle \sf = (2x + 1) \times 2$

$\displaystyle \sf = 4x + 2$

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