Question

A particle moves in a straight line in such a way that its velocity at any point is given by v^2 = 2-3x, where x is measured from a fixed point.The acceleration is

Answer 1

here is your answer OK

In this step follow you understand OK but not understand you see in this attachment OK

Dear Student,
Please find below the solution to the asked query:

Given above:

v2 = 2 - 3x ----- (1)

also we know that:
velocity = v = dx/dt
acceleration = a = dv/dt

and,
dividing and multiplying by dx on R.H.S.

dv/dt = dv/dx × dx/ dt ( and dx/dt = v)
a = dv/dt = v dv/dx ---- (2)

​thus,
a = dv/dx × dx/ dt

​differentiating (1) w.r.t. 'x'
2v dv/dx = d (2-3x) / dx
2v dv/dx = -3
v dv/dx = -3/2
comparing from equation (2) we have,
a = -3/2 correct answer....

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Answer 2

Given,

$v^{2} =2 - 3x \\ \\ \text{differentiating both sides with respect to x} \\ \\ 2v.\frac{dv}{dx} = 0 - 3 \\ \\ v \frac{dv}{dx} = \frac{-3}{2} \\ \\ \text{We know that,} \\ \\ v = \frac{dx}{dt} \\ \\ So, \\ \\ v \frac{dv}{dx} = \frac{-3}{2} \\ \\ \frac{dx}{dt} . \ \frac{dv}{dx} = \frac{-3}{2} \\ \\ \frac{dv}{dt} = \frac{-3}{2} \\ \\ a = \frac{-3}{2} unit/sec^{2} \ \ \ \ \text{(a =dv/dt)}$