Question

the speed v of a particle moving along a staright line when it is at a distance x from a fixed point on the line is given by v^2=108x-9x^2 then magnitude of its acceleration when it is at distance 3 meter from the fixed point​

Answer 1

Answer:

On differentiating with respect to time :

2.v. (dv/dt)=108.(dx/dt)-18.x.(dx/dt)

But a = dv/dt

v = dx/dt

So 2va = 108v - 18xv (equation 2)

Thus 2a = 108 - 18x

So if distance (x) = 3 :

2a = 108 - 54

a = 27

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Answer 2

Answer:

answer of this question is 27m/s^2....