91Ó°ÊÓ

Acceleration is a measure of how quickly a particle's velocity changes over time. In this exercise, the particle's acceleration depends on its position, given by the formula \( a = -k s^2 \), where \( k \) is a constant and \( s \) is the position. This kind of relationship suggests that the particle's acceleration changes with the square of the position. Such relationships are common in fields involving forces that vary with distance, like gravity or electrostatics.
In this scenario, the negative sign indicates that the acceleration is in the opposite direction to the increase in position, meaning it's acting to decelerate the particle. Solving for velocity here involves understanding how displacement affects acceleration, and then linking this with the particle's change in velocity over time.

In mathematics, a differential equation relates a function to its derivatives, and integrating such equations helps us find solutions that describe the behavior of a system. Here, our task is to find the velocity as a function of position.
The given acceleration equation \( a = -k s^2 \) can be transformed using the relationship \( a = v \frac{dv}{ds} \), enabling us to express the problem as a separable differential equation. The integration becomes:

• Move terms to separate each variable: \( v dv = -k s^2 ds \).
• Integrate both sides: \( \int v dv = \int -k s^2 ds \).
• After integration, these become \( \frac{v^2}{2} \) on the left and \( -\frac{ks^3}{3} + C \) for the right, where \( C \) is a constant.

These steps help determine how velocity changes with position, paving the way to solve for precise values, given initial conditions or other specific parameters.

Initial conditions are values set at the beginning of a problem that allow us to find unknown constants when solving differential equations. In this problem, the initial conditions give us specific values for position \( s_0 = 3 \text{ m} \) and velocity \( v_0 = 10 \text{ m/s} \).
When integrated, differential equations include arbitrary constants. Initial conditions help to determine these constants specifically. By plugging in the initial conditions into the integrated equation:

• You substitute \( s_0 \) and \( v_0 \) into the expression for velocity.
• Calculate the value of \( C \), ensuring the solution fits the real-world scenario at those starting values.

For this exercise, identifying and applying the initial conditions led to determining the integration constant \( C = 50 + 9k \). Subsequently, this value aids in evaluating the velocity function at other positions, such as at \( s = 5 \text{ m} \), to obtain the final desired result.