If the velocity of a particle is given by v=√(180-16x).then the acce

1.	If the velocity of a particle is given by v=√(180-16x).then the acceralation will be
Answer» We KNOW that acceleration is rate of CHANGE of velocity. We have, v=(180–16x)^1/2 Now,differentiating it w.r.t t i.e time, So, dv/dt = [1/2(180–16x)^(-1/2)](-16dx/dt) (I HOPE u have no problem in differentiating, if u have then please recinsider because it's BASICS) So, on simplifying above equation, a(acc.)=dv/dt=[1/2(180–16x)^(-1/2)](-16v) Putting the VALUE of v from original equation, a=dv/dt=[-8(180–16x)^(-1/2)](180–16x)^1/2) Clearly,on simplifying we get, a=dv/dt=-8 m/sec^2 which is the required answer.

If the velocity of a particle is given by v=√(180-16x).then the acceralation will be

Answer»

We KNOW that acceleration is rate of CHANGE of velocity.

We have, v=(180–16x)^1/2

Now,differentiating it w.r.t t i.e time,

So, dv/dt = [1/2(180–16x)^(-1/2)](-16dx/dt)

(I HOPE u have no problem in differentiating, if u have then please recinsider because it's BASICS)

So, on simplifying above equation,

a(acc.)=dv/dt=[1/2(180–16x)^(-1/2)](-16v)

Putting the VALUE of v from original equation,

a=dv/dt=[-8(180–16x)^(-1/2)](180–16x)^1/2)

Clearly,on simplifying we get,

a=dv/dt=-8 m/sec^2 which is the required answer.

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