Understanding the Relationship Between Displacement and Uniform Acceleration
In physics, the terms displacement and uniform acceleration are often interrelated. The assertion states that when the displacement of a body is directly proportional to the square of the time, the body is moving with uniform acceleration. The reason supports this by explaining the relationship between the slope of a velocity-time graph and acceleration. However, the reason does not directly support the assertion, making the final answer 1 point.
Assertion: Displacement Proportional to Time Squared
The assertion is that when the displacement s of a body is directly proportional to the square of the time t, it can be expressed as:
[s kt^2]
where k is a constant. This relationship is characteristic of uniform acceleration. The equation of motion for uniformly accelerated linear motion is:
[s frac{1}{2}at^2]
Comparing the two equations, it is clear that the presence of the square term in the displacement equation indicates uniform acceleration. This is because the acceleration a is constant, which is a fundamental property of uniform acceleration.
Reason: Slope of Velocity-Time Graph
The reason states that the slope of the velocity-time graph with the time axis gives acceleration. This is true because acceleration is defined as the change in velocity over the change in time:
[a frac{Delta v}{Delta t}]
If the relationship between displacement and time is given by s kt^2, then the velocity (v) is the first derivative of displacement with respect to time:
[v frac{ds}{dt} 2kt]
And the acceleration (a) is the first derivative of velocity with respect to time:
[a frac{dv}{dt} 2k]
Therefore, the acceleration is a constant value, which is a key characteristic of uniform acceleration.
Double Derivative of Displacement
Another way to understand the relationship is through the double derivative of the displacement with respect to time. The double derivative of displacement with respect to time is equal to the acceleration:
[frac{d^2s}{dt^2} 2k]
Since the value of (2k) is constant, the acceleration is also constant. This confirms that the relationship between displacement and time squared implies uniform acceleration.
Example: Free Fall Motion
Consider the example of an object in free fall under gravity, initially at rest ((u0)). The displacement (y) as a function of time (t) is given by:
[y frac{1}{2}gt^2]
Here, (g) is the acceleration due to gravity, which is a constant. This equation is a specific case of the general relationship between displacement and time for uniform acceleration. The constant acceleration in this case is (g), demonstrating the relationship between displacement, velocity, and acceleration.
Conclusion
The assertion is correct because the displacement proportional to the square of time indicates uniform acceleration. The reason is also correct but it does not directly explain or support the assertion. Therefore, while both statements are true, there is no direct causative connection between them.
The Final Answer
1 point for the assertion being correct but the reason does not directly support the assertion. This clarifies that while both the assertion and the reason are individually correct, they do not provide a direct logical link without additional context or explanation.