Question

The velocity of a particle is given by the equation \( v(x) = 3x^2 - 4x \), where \( x \) is the distance covered by the particle. The expression for its acceleration is:

• A. \( (6x - 4) \)
• B. \( 6(3x^2 - 4x) \)
• C. \( (3x^2 - 4x) (6x - 4) \)
• D. \( (6x - 4)^2 \)

Hint:

When given the velocity function, differentiate it with respect to \( x \) to find the acceleration.

Answer 1

The Correct Option is C

The acceleration \( a(x) \) is the rate of change of velocity with respect to time, given by the formula: \[ a(x) = \frac{dv}{dt}. \] Using the chain rule, we express this as: \[ a(x) = v'(x) \cdot v(x), \] where \( v(x) = 3x^2 - 4x \). First, we differentiate \( v(x) \): \[ v'(x) = \frac{d}{dx}(3x^2 - 4x) = 6x - 4 \] Now, the acceleration is: \[ a(x) = (6x - 4) \cdot (3x^2 - 4x). \]

Answer 2

Given:
Velocity as a function of position:
\[ v(x) = 3x^2 - 4x \] Find the acceleration \( a \).

Step 1: Recall that acceleration as a function of position is:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \]

Step 2: Calculate \( \frac{dv}{dx} \):
\[ \frac{dv}{dx} = \frac{d}{dx} (3x^2 - 4x) = 6x - 4 \]

Step 3: Substitute into acceleration formula:
\[ a = v \frac{dv}{dx} = (3x^2 - 4x)(6x - 4) \]

Therefore, the expression for acceleration is:
\[ \boxed{(3x^2 - 4x)(6x - 4)} \]