Question

A particle moves along a straight line such that its position is given by \( x(t) = 3t^2 - t^3 \). What is its velocity at \( t = 2 \) seconds?

• A. \( 0 \text{ m/s} \)
• B. \( 4 \text{ m/s} \)
• C. \( -4 \text{ m/s} \)
• D. \( 12 \text{ m/s} \)

Hint:

Whenever position is given as a function of time, remember: \[ \text{Velocity} = \frac{dx}{dt} \] and \[ \text{Acceleration} = \frac{d^2x}{dt^2} \] Differentiate carefully term-by-term using power rule.

Answer 1

The Correct Option is A

Concept: Velocity is the rate of change of position with respect to time. If the position of a particle is given by \(x(t)\), then its velocity is obtained by differentiating the position function: \[ v(t)=\frac{dx}{dt} \]

Step 1: Differentiate the position function to obtain velocity.
The given position function is: \[ x(t)=3t^2-t^3 \] Differentiate with respect to \(t\): \[ v(t)=\frac{d}{dt}(3t^2-t^3) \] \[ v(t)=6t-3t^2 \]

Step 2: Substitute \(t=2\) seconds into the velocity expression.
\[ v(2)=6(2)-3(2)^2 \] \[ v(2)=12-3(4) \] \[ v(2)=12-12 \] \[ v(2)=0 \] Hence, the velocity of the particle at \(t=2\) seconds is: \[ \boxed{0 \text{ m/s}} \]