Question

find the tangential and normal components \(\left(a_{T}\right.\) and \(\left.a_{N}\right)\) of the acceleration vector at \(t .\) Then evaluate at \(t=t_{1} .\) $$ x(t)=1+3 t, y(t)=2-6 t ; t_{1}=2 $$

Short answer

Both the tangential and normal components of acceleration are 0 at any time, including at \(t = 2\).

Step-by-step solution

Calculate the Velocity Components

First, find the velocity components by differentiating the position functions for both \(x(t)\) and \(y(t)\). This gives the velocity vector \(\mathbf{v}(t) = (v_x(t), v_y(t))\).\[v_x(t) = \frac{d}{dt}(1+3t) = 3\]\[v_y(t) = \frac{d}{dt}(2-6t) = -6\]

Compute the Speed

The speed \(v\) is the magnitude of the velocity vector. Calculate it using the formula:\[v(t) = \sqrt{v_x(t)^2 + v_y(t)^2} = \sqrt{3^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}\]

Calculate the Acceleration Components

Differentiate the velocity expressions to find the acceleration components \(\mathbf{a}(t) = (a_x(t), a_y(t))\).\[a_x(t) = \frac{d}{dt}(3) = 0\]\[a_y(t) = \frac{d}{dt}(-6) = 0\]Thus, \(\mathbf{a}(t) = (0, 0)\).

Compute the Tangential Component of Acceleration

The tangential component of acceleration \(a_T\) is given by:\[a_T = \frac{d}{dt} \left( v(t) \right)\]Since the speed \(v\) is constant \(3\sqrt{5}\), its derivative is 0.Thus, \(a_T = 0\).

Find the Normal Component of Acceleration

The normal component of acceleration \(a_N\) is computed using:\[a_N = \sqrt{a^2 - a_T^2}\]Since both \(a\) (the magnitude of \(\mathbf{a}(t)\)) and \(a_T\) are 0, it follows:\[a_N = \sqrt{0^2 - 0^2} = 0\]

Evaluate at \(t = t_1\)

Since both \(a_T\) and \(a_N\) are constants, evaluating at \(t = t_1 = 2\) simply confirms:\(a_T(t_1) = 0\) and \(a_N(t_1) = 0\).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that helps us understand how a function changes. It determines the rate at which one quantity changes with respect to another.
The derivative of a function represents this rate of change.

• If you think about position over time, the derivative gives you the velocity.
• Velocity, which measures the change of position, is the first derivative of a position function.
• To find acceleration, differentiate the velocity function once more.

This process of calculating derivatives is vital for understanding motion, as it provides insights into how speed and direction change over time.

Velocity Vector

A velocity vector is a mathematical representation of an object's speed and direction at any given instant. For a moving object, the velocity vector \(\mathbf{v}(t)\) provides a snapshot of motion in both horizontal and vertical directions.

• It's expressed in components: \(v_x(t)\) for the horizontal axis, and \(v_y(t)\) for the vertical axis.
• To calculate these components, differentiate the position functions: \(x(t)\) and \(y(t)\).
• Using the differentiation results in the exercise, the velocity at any time \(t\) is \(\mathbf{v}(t) = (3, -6)\).

Knowing the velocity vector allows us to determine the direction of motion and calculate speed when examining movement in a plane.

Acceleration Vector

The acceleration vector provides detailed information about how an object's velocity is changing over time. It's derived by taking the derivative of the velocity vector.

• Similar to velocity, the acceleration vector has components: \(a_x(t)\) and \(a_y(t)\).
• Differentiating the velocity components found in the exercise gives the acceleration \(\mathbf{a}(t) = (0, 0)\).
• This means there are no changes in the velocity vector components.

In situations where speed is constant, as in the given exercise, the acceleration vector being zero indicates that the object travels in a straight line without speeding up or slowing down.

Constant Speed

When an object is moving with constant speed, its rate of motion remains unchanged over time. This constant speed is directly linked to the magnitude of the velocity vector.

• To compute the speed, calculate the magnitude of the velocity vector.
• In the exercise, the speed was found to be \(3\sqrt{5}\).
• A constant speed means that the derivative of speed, i.e., the tangential acceleration \(a_T\), is zero.

Reaching a constant speed implies that all changes in direction do not affect the magnitude of speed, explaining why the tangential component of acceleration is zero.