91Ó°ÊÓ

Understanding velocity and acceleration is crucial in physics, especially when analyzing how objects move. Velocity is a vector quantity that describes the rate at which an object changes its position, incorporating both speed and direction.
In the exercise, the velocity of the toy airplane is given as a function of time, meaning it changes with time. The velocity \( \vec{v}(t) = (1.20t) \hat{\imath} + (12.0 - 2.00t) \hat{\boldsymbol{J}} \, \mathrm{m/s} \) specifically shows how the airplane's movement adjusts in the horizontal and vertical directions.
Acceleration, on the other hand, indicates how much the velocity changes over time. It is the derivative of the velocity function with respect to time, giving us \( \vec{a}(t) = (1.20) \hat{\imath} - (2.00) \hat{\boldsymbol{J}} \).

• Velocity relates to the position change.
• Acceleration details how this change in position is becoming faster or slower.

The dot product, also known as the scalar product, is a crucial operation in vector calculus. It takes two vectors and returns a single scalar (a number), providing important information about the vectors' orientation relative to each other.
For two vectors to be perpendicular (or orthogonal), their dot product must equal zero. In our problem, the task is to find the value of time \( t \) when the velocity vector is perpendicular to the acceleration vector, meaning their dot product is zero.
This requirement is implemented as follows: \( \vec{v}(t) \cdot \vec{a}(t) = (1.20t)(1.20) + (12.0 - 2.00t)(-2.00) = 0 \). Simplifying this expression helps us find when the vectors meet this condition.

• Dot product yields information on the angles between vectors.
• Perpendicular vectors have a dot product of zero.

Derivatives are fundamental in calculus and act as the foundation for understanding changes in quantities. They describe the rate at which one variable changes in relation to another.
In the context of this problem, the derivative of the velocity with respect to time gives us the acceleration. This allows us to understand how the velocity changes over time. Taking the derivative of \( \vec{v}(t) = (1.20t) \hat{\imath} + (12.0 - 2.00t) \hat{\boldsymbol{J}} \) results in the acceleration vector \( \vec{a}(t) = 1.20 \hat{\imath} - 2.00 \hat{\boldsymbol{J}} \. \)

• Velocity's derivative with respect to time is acceleration.
• Derivatives help in finding rates of change.

By differentiating, we quantify how swiftly velocity changes, which is key for dynamic systems.

In vector calculus, perpendicular vectors are fundamental due to their unique properties. Vectors are perpendicular when they intersect at right angles, implying that their dot product equals zero.
This orthogonality is significant when analyzing velocity and acceleration. For our airplane, at a specific time \( t \), the velocity is perpendicular to the acceleration if their dot product is zero. This provides a simple yet powerful condition used in the solution process: \( \vec{v}(t) \cdot \vec{a}(t) = 0 \).

• Perpendicular vectors indicate unique directional properties.
• When examining motion, it shows non-aligning forces or motions.

Understanding when vectors are perpendicular helps in applications like force balancing and analyzing motion trajectories.