91Ó°ÊÓ

To understand how to calculate the velocity of a particle in curvilinear motion, we begin by looking at the velocity components. The velocity vector, \( \mathbf{v} \), is derived from the position functions \( x(t) \) and \( y(t) \) by taking their time derivatives. These derivatives give us the velocity components along each axis: \( v_x(t) = \frac{dx}{dt} \) and \( v_y(t) = \frac{dy}{dt} \).

The position functions given are \( x(t) = 10.25t + 1.75t^2 - 0.45t^3 \) and \( y(t) = 6.32 + 14.65t - 2.48t^2 \). We differentiate these to find:
- For the x-component: \( v_x(t) = 10.25 + 3.5t - 1.35t^2 \)
- For the y-component: \( v_y(t) = 14.65 - 4.96t \)

At a specified time, like \( t = 5 \) s, you insert this value into both velocity components. You get \( v_x(5) = -6.0 \text{ mm/s} \) and \( v_y(5) = -10.15 \text{ mm/s} \).

To find the magnitude of the total velocity, we calculate the resultant velocity vector using:

• \( \| \mathbf{v} \| = \sqrt{v_x(5)^2 + v_y(5)^2} \)
• This results in \( \sqrt{36 + 103.0225} \approx 11.79 \text{ mm/s} \).

Understanding these steps is key to mastering velocity calculations in curvilinear motion.

The next step in analyzing curvilinear motion involves determining the acceleration components. These are obtained by differentiating the velocity components with respect to time. This second derivative of the position function provides us the acceleration along each axis.

The acceleration along the x-axis is found by differentiating \( v_x(t) = 10.25 + 3.5t - 1.35t^2 \):
- Yields: \( a_x(t) = 3.5 - 2.7t \)

For the y-axis, the acceleration is derived from \( v_y(t) = 14.65 - 4.96t \):
- Resulting in: \( a_y(t) = -4.96 \)

When \( t = 5 \) s, substituting into the acceleration components gives:
- \( a_x(5) = -10.0 \text{ mm/s}^2 \)
- \( a_y(5) = -4.96 \text{ mm/s}^2 \)

Understanding acceleration in curvilinear motion allows you to predict how the particle accelerates or decelerates over time, revealing much about its dynamic behavior.

Curvilinear motion problems often require solving quadratic equations, especially when determining the time at which certain conditions occur. In this case, one condition is when the velocity vector makes a \(45^\circ\) angle with the \(x\)-axis. The condition for this is \( \tan(\theta) = \frac{v_y}{v_x} = 1 \), meaning \( v_y = v_x \).

The problem simplifies to solving the equation:
- \( 14.65 - 4.96t = 10.25 + 3.5t - 1.35t^2 \)

Rearranging, we arrive at:
- \( 0 = 1.35t^2 - 8.46t - 4.4 \)

This equation is quadratic in form \( ax^2 + bx + c = 0 \). Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with:

• \( a = 1.35 \)
• \( b = -8.46 \)
• \( c = -4.4 \)

The calculation yields:
- \( t \approx 7.48 \) s

It's crucial for students to appreciate the math behind quadratic equations, as they frequently appear in physics, allowing for the calculation of critical times during motion.