Question

The graph shows the variation of v^2 v/s x. The magnitude of maximum acceleration is "x" m/s^2. Value of "x" is.


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Answer 1

Answer:

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Answer 2

x = 6m/s²

We know acceleration a = $v\frac{dv}{dx}$

Slope of this graph = $\frac{d(v^{2} )}{dx}$

                                = $2v\frac{dv}{dx}$

                                = $2a$

⇒ a = $\frac{Slope}{2}$ ............(A)

⇒for a to be maximum, slope(steepness) of given graph must be maximum

Clearly,slope is maximum in the interval x=2 to x=3

and is equal to $\frac{16-4}{3-2}$

                       = 12

from equation (A),

a = 12/2

  =6 m/s²