91Ó°ÊÓ

Acceleration tells us how quickly an object's velocity changes with time. It's valuable in examining dynamic systems and is often represented mathematically as the derivative of velocity with respect to time. In mathematical terms, acceleration \(a\) is given by: \[ a = \frac{dv}{dt} \]In our exercise, acceleration is expressed in terms of velocity \(v\) by \(a = 2 \sqrt{v}\). This simplifies understanding for problems where acceleration depends on velocity, rather than time. Such relationships are solved using differential equations. Differential equations, like the one in the exercise, express rates of change and typically require integration to find the velocity or position function. Properly solving this equation will yield the velocity function, making it crucial to understand how to handle this type of integration. By understanding acceleration, you can grasp how systems evolve over time. It gives insight into how external forces influence motion, making it a fundamental concept in physics.

Velocity is the rate at which an object changes its position. It is a vector property, having both magnitude and direction. In simpler terms, it tells us how fast something is moving and in which direction. In our exercise, velocity is tied directly to acceleration through the differential equation \(\frac{dv}{dt} = 2\sqrt{v}\). This signifies that as the velocity changes, so does acceleration, suggesting a dynamic and responsive motion.Solving for velocity when given acceleration involves integrating the differential equation. In our solution, integrating both sides provided a functional description of velocity over time, captured as \(v(t) = \left(\frac{2t + 4}{2}\right)^2\). This powerful equation enables us to calculate velocity at any given time or instant that we are interested in. Understanding velocity’s calculation through integration not only aids in understanding motion over time, but also its change — a fundamental concept in kinematics and beyond.

The integration constant, often symbolized as \(C\), appears when solving differential equations because integration is the reverse process of differentiation, which 'loses' a constant during differentiation. After integrating the differential equation \(\frac{dv}{dt} = 2\sqrt{v}\), we arrived at the equation \(2\sqrt{v} = 2t + C\). Here, \(C\) accounts for the initial conditions of our problem. To determine \(C\), we utilize initial conditions given in the problem statement. In this case, at time \(t=2\text{s}\), the velocity \(v=16\text{ms}^{-1}\). Substituting these values in the integrated equation, we solve for the constant \(C\). It bridges general solutions to specific scenarios described by the problem.Without calculating \(C\), the solution would lack specificity and precision. The correct value of \(C\) ensures that the solution accurately reflects the system's behavior, ensuring all computations align correctly with the described motion.