91Ó°ÊÓ

In physics, acceleration is the rate of change of velocity over time. It's a vector quantity, meaning it has both magnitude and direction. If an object starts from rest, like our car with an initial velocity of zero, it gains speed due to acceleration. The acceleration equation can be written as:

• The velocity at any time, assumed to be starting from a stationary state, is given by: \( v = at \) where \( v \) is the final velocity, \( a \) is the acceleration, and \( t \) is the time.
• The position or distance travelled is defined by another kinematic equation: \( s = ut + \frac{1}{2}at^2 \) Again with \( u = 0 \), this simplifies to \( s = \frac{1}{2}at^2 \).

These equations are crucial for understanding how an object in motion behaves under constant acceleration, especially for determining changes in kinetic energy over time. They help in linking the physical quantity of acceleration to how an object's movement evolves.

The work-energy theorem is a vital principle that connects the concepts of work and kinetic energy. It states that the work done on an object is equal to the change in its kinetic energy. This theorem can be expressed as: \[ W = \Delta K \] Where \( W \) is the work done and \( \Delta K \) is the change in kinetic energy. For our exercise:

• The initial kinetic energy \( K_0 = 0 \) since the car starts from rest.
• At time \( t \), the kinetic energy is: \[ K = \frac{1}{2}ma^2t^2 \]
• Therefore, the work done from \( t = 0 \) to time \( t \) is: \[ W = \frac{1}{2}ma^2t^2 \]

This means that all the energy transferred to the car in the form of work translates directly into kinetic energy, accounting for the speeding up of the car under constant acceleration.

Power indicates how quickly work is being done or energy is being transferred. It is defined as the rate of doing work or the rate at which energy changes form. The formula relating power and work is: \[ P = \frac{dW}{dt} \] Where \( P \) is power, \( W \) is work, and \( dt \) denotes time differentiation. In the context of our car exercise:

• Given: \( W = \frac{1}{2}ma^2t^2 \)
• The derivative with respect to time gives us: \[ P = ma^2t \]

This tells us how much power the car expends at any moment in time as it accelerates. If two cars accelerate with different rates, the faster accelerating car, at any instance, expends more power. As in our example, a car with twice the acceleration (\( 2a \)) will use four times (4) the power compared to a car with acceleration \( a \). This significant increase shows the strong dependency of power on acceleration.