If the velocity of a particle is given by υ = (180 – 16x)1/2 m/s, then its acceleration will be :Zero8 m/s2– 8 m/s24 m/s2

The acceleration of a particle is given by the derivative of its velocity with respect to time.

Given the velocity function υ = (180 – 16x)^(1/2), we first need to find its derivative with respect to x, which will give us the rate of change of velocity, or acceleration.

The derivative of a function f(x) = (u(x))^(1/2) is given by f'(x) = (1/2) * u'(x) / (u(x))^(1/2), where u'(x) is the derivative of u(x) with respect to x.

Here, u(x) = 180 - 16x. So, u'(x) = -16.

Substituting these values into the formula, we get:

a = f'(x) = (1/2) * (-16) / ((180 - 16x)^(1/2)) = -8 / ((180 - 16x)^(1/2))

So, the acceleration of the particle is -8 / ((180 - 16x)^(1/2)) m/s^2.

Please note that this is the acceleration as a function of x. The actual acceleration at a specific point in time will depend on the value of x at that time.