The velocity of a particle moving along x-axis is given as `v=x^(2)-5x+4` (in m`//`s) where x denotes the x-coordinate of the particle in metres. Find the magnitude of acceleration of the particle when the velocity of particle is zero?

Options

A. `0 m//s^(2)`
B. `2 m//s^(2)`
C. `3 m//s^(2)`
D. None of these

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Set the velocity equation to zero
The velocity of the particle is given by the equation:
\[ v = x^2 - 5x + 4 \]

To find when the velocity is zero, we set the equation to zero:
\[ x^2 - 5x + 4 = 0 \]

### Step 2: Factor the quadratic equation
Next, we will factor the quadratic equation:
\[ x^2 - 5x + 4 = (x - 1)(x - 4) = 0 \]

### Step 3: Solve for x
Setting each factor to zero gives us:
1. \( x - 1 = 0 \) → \( x = 1 \)
2. \( x - 4 = 0 \) → \( x = 4 \)

Thus, the particle has zero velocity at \( x = 1 \) m and \( x = 4 \) m.

### Step 4: Find acceleration
Acceleration \( a \) can be expressed as:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \cdot \frac{dv}{dx} \]

### Step 5: Calculate \( \frac{dv}{dx} \)
First, we need to find \( \frac{dv}{dx} \):
\[ v = x^2 - 5x + 4 \]
Taking the derivative:
\[ \frac{dv}{dx} = 2x - 5 \]

### Step 6: Evaluate \( \frac{dv}{dx} \) at \( x = 1 \) and \( x = 4 \)
1. For \( x = 1 \):
\[ \frac{dv}{dx} = 2(1) - 5 = 2 - 5 = -3 \]

2. For \( x = 4 \):
\[ \frac{dv}{dx} = 2(4) - 5 = 8 - 5 = 3 \]

### Step 7: Calculate acceleration at both points
Now we can calculate the acceleration at both points where \( v = 0 \):
1. At \( x = 1 \):
\[ a = v \cdot \frac{dv}{dx} = 0 \cdot (-3) = 0 \, \text{m/s}^2 \]

2. At \( x = 4 \):
\[ a = v \cdot \frac{dv}{dx} = 0 \cdot 3 = 0 \, \text{m/s}^2 \]

### Conclusion
The magnitude of acceleration of the particle when the velocity is zero is:
\[ \text{Magnitude of acceleration} = 0 \, \text{m/s}^2 \]

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