91Ó°ÊÓ

To analyze the motion of an object along a curve, we often use parametrization to express the object's position as a function of a parameter, typically time. In the given exercise, the object's path is described by the parabola equation \( y = x^2 \). To explore this motion further, we introduce a parameter \( t \), setting \( x(t) = t \) and \( y(t) = t^2 \).

This approach allows us to write the position of the object in vector form: \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} = t \mathbf{i} + t^2 \mathbf{j} \). By representing curves parametrically, we can easily calculate derivatives such as velocity and acceleration, essential for dynamic analysis. This method provides a systematic way to understand how an object moves along complex paths by breaking them down into simpler components.

The velocity vector describes how an object's position changes over time. It is calculated as the derivative of the position vector with respect to time. In this case, the position vector is \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \).

Differentiating \( \mathbf{r}(t) \) with respect to \( t \) gives us the velocity vector: \( \mathbf{v}(t) = \frac{d}{dt}[t \mathbf{i} + t^2 \mathbf{j}] = \mathbf{i} + 2t \mathbf{j} \). This vector indicates both the direction and magnitude of the object's velocity at any point along its path.

To satisfy the condition of constant speed, we ensure that the magnitude of the velocity vector remains unchanged. Given that our constant speed is 10 units/s, we find: \( |\mathbf{v}| = \sqrt{1 + 4t^2} = 10 \). Solving this equation lets us determine appropriate values for \( t \).

The acceleration vector reflects changes in the velocity of an object over time. To find it, we take the derivative of the velocity vector. From our earlier calculation, we have the velocity vector \( \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} \).

Thus, the acceleration vector is found by differentiating the velocity vector: \( \mathbf{a}(t) = \frac{d}{dt}(\mathbf{i} + 2t \mathbf{j}) = 2 \mathbf{j} \). Unlike velocity, acceleration does not have a component in the \( \mathbf{i} \) direction here, indicating that all changes in velocity are vertically aligned, according to our parabola path.

Acceleration is crucial for understanding force because it directly ties into Newton's Second Law, making it essential for calculating net forces acting on the object.

Newton's Second Law is a cornerstone of classical mechanics, stating that the force exerted on an object is equal to the mass of the object multiplied by its acceleration: \( \mathbf{F} = m \mathbf{a} \). This law connects the object's motion with the forces acting upon it and is vital for analyzing dynamic systems.

In our problem, the acceleration vector is \( 2 \mathbf{j} \). Using Newton's Second Law, we calculate the force exerted on the object at the given point \((\sqrt{2}, 2)\): \( \mathbf{F} = m \cdot 2 \mathbf{j} = 2m \mathbf{j} \).

This simplification shows that the force acting on the object is purely vertical, aligned with the acceleration vector. Understanding this relationship is key to solving physics problems involving motion along curves.