If the velocity of a particle is given by `v=(180-16x)^((1)/(2))(m)/(s)`, then its acceleration will be

Options

A. zero
B. ` 8 m//s^(2) `
C. ` -8m//s ^(2) `
D. ` 4 m//s^(2) `

Explanation

To find the acceleration of the particle given the velocity equation \( v = (180 - 16x)^{1/2} \) m/s, we can follow these steps:

### Step 1: Understand the relationship between velocity and acceleration
Acceleration \( a \) can be expressed as the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Using the chain rule, we can express this as:
\[ a = \frac{dv}{dx} \cdot \frac{dx}{dt} \]
Since \( \frac{dx}{dt} = v \), we have:
\[ a = v \cdot \frac{dv}{dx} \]

### Step 2: Differentiate the velocity function
We need to find \( \frac{dv}{dx} \). Start with the given velocity equation:
\[ v = (180 - 16x)^{1/2} \]
To differentiate \( v \) with respect to \( x \), we can use the chain rule:
\[ \frac{dv}{dx} = \frac{1}{2}(180 - 16x)^{-1/2} \cdot (-16) \]
This simplifies to:
\[ \frac{dv}{dx} = -\frac{8}{(180 - 16x)^{1/2}} \]

### Step 3: Substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration formula
Now substitute \( v \) and \( \frac{dv}{dx} \) into the acceleration equation:
\[ a = v \cdot \frac{dv}{dx} \]
Substituting the expressions we found:
\[ a = (180 - 16x)^{1/2} \cdot \left(-\frac{8}{(180 - 16x)^{1/2}}\right) \]
The \( (180 - 16x)^{1/2} \) terms cancel out:
\[ a = -8 \text{ m/s}^2 \]

### Final Answer
Thus, the acceleration of the particle is:
\[ \boxed{-8 \text{ m/s}^2} \]

---