If the displacement of a car is proportional to the square of time, then the car is moving with

Analysis

The question asks us to determine the type of motion a car exhibits when its displacement ($s$) is directly proportional to the square of time ($t^2$).

Concept

In kinematics, we use the equations of motion to describe how displacement, velocity, and acceleration relate to time.

• Displacement ($s$): The distance moved in a specific direction.
• Velocity ($v$): The rate of change of displacement with respect to time ($v = \frac{ds}{dt}$).
• Acceleration ($a$): The rate of change of velocity with respect to time ($a = \frac{dv}{dt}$).

Step-by-Step Solution

1. Translate the statement into a mathematical equation: We are told that displacement ($s$) is proportional to the square of time ($t^2$): $$s \propto t^2$$This can be written as:$$s = kt^2$$(where$k$ is a constant of proportionality).

2. Find the velocity ($v$): Velocity is the first derivative of displacement with respect to time: $$v = \frac{ds}{dt} = \frac{d}{dt}(kt^2)$$ $$v = 2kt$$

3. Find the acceleration ($a$): Acceleration is the derivative of velocity with respect to time: $$a = \frac{dv}{dt} = \frac{d}{dt}(2kt)$$ $$a = 2k$$

4. Interpret the result: Since $k$ is a constant, $2k$ is also a constant. This means the acceleration of the car does not change over time. When acceleration is constant, we describe the motion as uniform acceleration.

Conclusion

Since the acceleration is constant ($a = 2k$), the car is moving with uniform acceleration.

Exam Tip: Remember this simple hierarchy for motion:

• If $s \propto t^0$ (constant), velocity is zero.
• If $s \propto t^1$ (linear), velocity is constant (uniform velocity).
• If $s \propto t^2$ (quadratic), acceleration is constant (uniform acceleration).