To analyze the motion of a particle with an initial velocity \( u \) and a retardation proportional to its velocity \( av \), we can use the concepts of kinematics and calculus. The retardation means that the particle is slowing down, and we can express this mathematically. Let's break down the problem step by step to understand the implications of the given conditions.
Understanding Retardation
Retardation is a type of acceleration that reduces the velocity of an object. In this case, the retardation is given as \( av \), where \( a \) is a constant and \( v \) is the velocity at any time \( t \). This means that the rate of change of velocity is proportional to the current velocity, which is characteristic of exponential decay.
Mathematical Representation
The equation of motion can be expressed as:
- \( \frac{dv}{dt} = -av \)
This differential equation indicates that the velocity decreases over time. To solve it, we can separate variables:
- \( \frac{dv}{v} = -a dt \)
Integrating both sides gives us:
Exponentiating both sides results in:
- \( v = e^{-at + C} = e^C e^{-at} \)
Letting \( e^C = u \) (the initial velocity), we find:
Distance Covered by the Particle
To find the distance covered by the particle, we can integrate the velocity function:
- Distance \( s = \int v(t) dt = \int u e^{-at} dt \)
The integral of \( e^{-at} \) is \( -\frac{1}{a} e^{-at} \), so:
- \( s = -\frac{u}{a} e^{-at} + C \)
At \( t = 0 \), the distance \( s = 0 \), which gives us \( C = \frac{u}{a} \). Thus, the distance covered becomes:
- \( s(t) = \frac{u}{a} (1 - e^{-at}) \)
Key Insights
Now, let's evaluate the options given in your question:
- (a) The particle will cover a total distance \( \frac{u}{a} \): This is true as \( t \) approaches infinity, \( e^{-at} \) approaches 0, and \( s \) approaches \( \frac{u}{a} \).
- (b) The particle will come to rest after \( \frac{1}{a} \): This is misleading. The particle never truly comes to rest in a finite time; it asymptotically approaches zero velocity.
- (c) The particle will continue to move for a very long time: This is accurate. The particle will keep moving but will slow down indefinitely.
- (d) The velocity of the particle will become \( \frac{u}{2} \) after time \( \frac{1}{a} \): This can be calculated. At \( t = \frac{1}{a} \), \( v = u e^{-1} \), which is not equal to \( \frac{u}{2} \) unless \( e \) is approximately 2, which it is not.
Final Thoughts
In summary, the correct statements are (a) and (c). The particle will asymptotically approach a total distance of \( \frac{u}{a} \) and will continue to move indefinitely, albeit at a decreasing speed. Understanding these principles helps in grasping the nature of motion under variable acceleration conditions.
