91Ó°ÊÓ

Velocity is a measure of how fast something is moving in a specific direction. In kinematics, calculating velocity involves determining the rate of change of an object's position. This can be done by differentiating the position with respect to time.

For the problem at hand, we have a particle moving along a curve given by the equation:

• \( x = \frac{t^3}{3} \)
• \( y = \frac{x}{2} \)

To find the velocity, we need to compute the first derivative of these position functions. For \( x \), differentiate to get \( \dot{x} = t^2 \).

Next, substitute \( x = \frac{t^3}{3} \) into the equation for \( y \), resulting in \( y = \frac{t^3}{6} \). Differentiating \( y \) with respect to \( t \) gives \( \dot{y} = \frac{t^2}{2} \).

To find the velocity at a specific time, plug \( t = 1 \text{ s} \) into the equations. We find \( \dot{x} = 1 \) and \( \dot{y} = \frac{1}{2} \). Thus, the velocity vector at that moment is \( \hat{i} + \frac{1}{2} \hat{j} \), combining the horizontal and vertical components.

Acceleration tells us how quickly an object is speeding up or slowing down, and it also involves a specific direction. It is the rate of change of velocity with respect to time. To find acceleration, we must differentiate the velocity functions.

Starting with the velocity components obtained from the previous section:

• \( \dot{x} = t^2 \)
• \( \dot{y} = \frac{t^2}{2} \)

Differentiate these to find the acceleration. For \( \dot{x} \), the derivative is \( \ddot{x} = 2t \). For \( \dot{y} \), it's \( \ddot{y} = t \).

The acceleration at \( t = 2 \text{ s} \) becomes \( \ddot{x} = 4 \) and \( \ddot{y} = 2 \). Therefore, the acceleration vector is \( 4 \hat{i} + 2 \hat{j} \). This result mismatches the given options, which could lead to revisiting the exercise assumptions or the provided answer choices.

Differentiation is a crucial mathematical tool in physics, particularly in kinematics. It allows us to find the rate at which quantities change with respect to time. This is essential for understanding concepts like velocity and acceleration.

In physics problems, the relation between position, velocity, and acceleration is often explored through differentiation:

• The first derivative of position with respect to time gives velocity. This tells you how fast and in what direction an object moves.
• The second derivative of position, or the first derivative of velocity, gives acceleration. This measures changes in the rate of motion.

In our exercise, differentiating the position equations allowed us to find these important rates:

• First, we differentiated the equations for position (\( x \) and \( y \)) to get velocity.
• Then, we further differentiated the results to find acceleration.

This method is universally applicable in problems involving motion, making differentiation one of the key skills in physics.