If the velocity of a car is given by `V=(150-10x)^(1//2)` m/s . If car retards their motion by applying brakes then what will be the acceleration ?

Options

A. ` 1"m/s"^(2)`
B. `2 "m/s"^(2)`
C. `5"m/s"^(2)`
D. `-5"m/s"^(2)`

Explanation

To find the acceleration of the car when it is retarding its motion, we start with the given velocity function:

\[ V = (150 - 10x)^{1/2} \, \text{m/s} \]

### Step 1: Differentiate the velocity with respect to time
Acceleration \( a \) is defined as the rate of change of velocity with respect to time, which can be expressed as:

\[ a = \frac{dV}{dt} \]

Using the chain rule, we can express this as:

\[ a = \frac{dV}{dx} \cdot \frac{dx}{dt} \]

Where \( \frac{dx}{dt} \) is the velocity \( V \).

### Step 2: Differentiate \( V \) with respect to \( x \)
To find \( \frac{dV}{dx} \), we differentiate the velocity function:

\[ V = (150 - 10x)^{1/2} \]

Using the power rule and chain rule, we get:

\[ \frac{dV}{dx} = \frac{1}{2}(150 - 10x)^{-1/2} \cdot (-10) \]

This simplifies to:

\[ \frac{dV}{dx} = -\frac{5}{(150 - 10x)^{1/2}} \]

### Step 3: Substitute \( \frac{dx}{dt} \) with \( V \)
Now, substituting \( \frac{dx}{dt} = V \) into the expression for acceleration:

\[ a = \frac{dV}{dx} \cdot V \]

Substituting \( \frac{dV}{dx} \):

\[ a = -\frac{5}{(150 - 10x)^{1/2}} \cdot V \]

### Step 4: Substitute \( V \) back into the equation
Since \( V = (150 - 10x)^{1/2} \), we can substitute this back into the equation for acceleration:

\[ a = -\frac{5}{(150 - 10x)^{1/2}} \cdot (150 - 10x)^{1/2} \]

This simplifies to:

\[ a = -5 \, \text{m/s}^2 \]

### Conclusion
Thus, the acceleration of the car when it applies brakes is:

\[ a = -5 \, \text{m/s}^2 \]