Understanding Motion: Displacement and Time Relationships
In physics, the relationship between displacement ($s$) and time ($t$) tells us everything we need to know about an object's state of motion. When we are given a relationship like $s \propto t^2$, we can use calculus to determine the velocity and acceleration of the object.
The Mathematical Derivation
Define the relationship: We are given that displacement ($s$) is proportional to the square of time ($t^2$): $s = kt^2$ (where $k$ is a constant of proportionality)
Find Velocity ($v$): Velocity is defined as the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(kt^2) = 2kt$ Since velocity is proportional to time ($v \propto t$), the velocity is changing over time. This means the object is not moving with uniform velocity.
Find Acceleration ($a$): Acceleration is defined as the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(2kt) = 2k$ Since $k$ is a constant, $2k$ is also a constant. Therefore, the acceleration is constant over time.
Conclusion
Because the acceleration remains constant ($2k$), the body is moving with uniform acceleration.
Intuition
- Uniform Velocity: This would mean displacement is simply a linear function of time ($s = vt$). If you graphed this, you would get a straight line.
- Uniform Acceleration: This happens when velocity changes at a steady rate. As seen in the kinematic equation $s = ut + \frac{1}{2}at^2$, if initial velocity $u$ is zero, then $s = \frac{1}{2}at^2$, which confirms that $s \propto t^2$ is the signature of constant acceleration.