91Ó°ÊÓ

When we talk about displacement-time relationships in physics, it's all about understanding how the position of an object changes over time. Here, the given displacement equation is \( s = \frac{1}{3}t^2 \). This means the displacement \( s \) depends on the square of time \( t \).
In simpler terms, if you graph this equation with displacement on the y-axis and time on the x-axis, you will get a parabolic curve opening upwards. This indicates that the further out in time you get, the more the displacement increases.

• Time \( t \) is the independent variable, meaning it's what you choose or change, like setting a stopwatch.
• Displacement \( s \) is the dependent variable, meaning it changes in response to changing time.
• The coefficient \( \frac{1}{3} \) gives the rate at which displacement increases over time.

Understanding this relationship helps us predict how far an object moves in a given time, which is crucial in solving physics problems.

Velocity tells us how fast something is moving and in which direction. To find the velocity from a displacement-time relationship, we differentiate the displacement equation with respect to time. In this case, the displacement equation is \( s = \frac{1}{3}t^2 \).
Differentiating gives us \( v(t) = \frac{d}{dt}(\frac{1}{3}t^2) = \frac{2}{3}t \). This formula means that the velocity \( v \) increases linearly with time.

• At any specific moment, substitute the time value into the velocity equation to find the speed.
• The velocity here is expressed in meters per second (m/s), showing rate and direction.
• For \( t = 2 \) seconds, the velocity is \( \frac{4}{3} \text{ m/s} \).

Velocity is crucial because it helps determine how quickly an object will reach a certain point.

Acceleration indicates how the object’s velocity changes with time. In our exercise, acceleration is found by differentiating the velocity function \( v(t) = \frac{2}{3}t \) with respect to time, resulting in a constant acceleration \( a(t) = \frac{2}{3} \text{ m/s}^2 \). This means that the object gains \( \frac{2}{3} \) meters per second of speed every second, which is steady due to the constant force.
According to Newton’s Second Law, \( F = ma \), acceleration helps us find the force needed to change the motion of the body. Here,

• The mass \( m \) is \( 3 \text{ kg} \).
• The acceleration \( a \) is \( \frac{2}{3} \text{ m/s}^2 \).

Plug these into Newton's Second Law to find the force: \( F = 3 \times \frac{2}{3} = 2 \text{ N} \). This law shows the direct proportional relationship between force and the product of mass and acceleration.

Work in physics refers to the energy transferred to or from an object via the application of a force along a displacement. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy.
The formula for work done \( W \) is \( W = Fs \). Here, the force \( F \) is \( 2 \text{ N} \) and the displacement \( s \) over 2 seconds is \( \frac{4}{3} \text{ meters} \). Therefore, the work done is:
\[ W = 2 \times \frac{4}{3} = \frac{8}{3} \text{ J} \]

• Work is calculated in joules \( \text{J} \).
• It considers only the component of the force in the direction of displacement.
• The calculated work indicates the amount of energy transferred as a result of the force applied over \( \frac{4}{3} \) meters.

This principle is fundamental in physics, linking the concepts of force, energy, and motion.