A particle moving with uniform acceleration has its final velocity `(v=sqrt(150+8x))` m/s where x is the distance travelled by the body. Then the acceleration is

Options

A. `+4 "m/s"^2`
B. `-4 "m/s"^2`
C. `+8 "m/s"^2`
D. `-8 "m/s"^2`

Answer

To find the acceleration of a particle moving with uniform acceleration, given the final velocity \( v = \sqrt{150 + 8x} \), we can follow these steps:

### Step 1: Square the Velocity Equation
Start with the given equation for final velocity:
\[ v = \sqrt{150 + 8x} \]
Square both sides to eliminate the square root:
\[ v^2 = 150 + 8x \]

### Step 2: Differentiate with Respect to Time
Differentiate both sides of the equation with respect to time \( t \):
\[ \frac{d(v^2)}{dt} = \frac{d(150 + 8x)}{dt} \]
Using the chain rule on the left side:
\[ 2v \frac{dv}{dt} = 0 + 8 \frac{dx}{dt} \]
Since \( \frac{dx}{dt} = v \), we can substitute this into the equation:
\[ 2v \frac{dv}{dt} = 8v \]

### Step 3: Simplify the Equation
Assuming \( v \neq 0 \) (since we are considering a moving particle), we can divide both sides by \( v \):
\[ 2 \frac{dv}{dt} = 8 \]

### Step 4: Solve for Acceleration
Now, isolate \( \frac{dv}{dt} \):
\[ \frac{dv}{dt} = 4 \]
This represents the acceleration \( a \):
\[ a = 4 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the particle is:
\[ \boxed{4 \, \text{m/s}^2} \]

---