Finding the Components of Velocity and Acceleration in a Given Direction

This article will guide you through the process of determining the components of a particle's velocity and acceleration at a specific point in time, in a particular direction. We will use a detailed example to illustrate the method.

Introduction

When dealing with vector calculus, understanding how to find velocity and acceleration vectors for a moving particle in 3-dimensional space is fundamental. This article provides a step-by-step guide to finding the velocity and acceleration vectors at a given time, and further breaks down these vectors into components in a specified direction.

Position Vector and Derivatives

The position of a particle moving along a curve can be described by a position vector where . For this example, the position vector is given by .

Step 1: Position Vector

The position of the particle is given by:

[ x(t) 2t^2, quad y(t) t^2 - 4t, quad z(t) 3t - 5]

We can express this in vector form:

[ mathbf{r}(t) langle 2t^2, t^2 - 4t, 3t - 5 rangle]

Step 2: Velocity Vector

The velocity vector is the derivative of the position vector with respect to time :

[ mathbf{v}(t) frac{dmathbf{r}}{dt} langle frac{dx}{dt}, frac{dy}{dt}, frac{dz}{dt} rangle]

Calculating each component:

[ frac{dx}{dt} frac{d}{dt}(2t^2) 4t] [ frac{dy}{dt} frac{d}{dt}(t^2 - 4t) 2t - 4] [ frac{dz}{dt} frac{d}{dt}(3t - 5) 3]

Thus, the velocity vector is:

[ mathbf{v}(t) langle 4t, 2t - 4, 3 rangle]

Step 3: Acceleration Vector

The acceleration vector is the derivative of the velocity vector with respect to time :

[ mathbf{a}(t) frac{dmathbf{v}}{dt} langle frac{d^2x}{dt^2}, frac{d^2y}{dt^2}, frac{d^2z}{dt^2} rangle]

Calculating each component:

[ frac{d^2x}{dt^2} frac{d}{dt}(4t) 4] [ frac{d^2y}{dt^2} frac{d}{dt}(2t - 4) 2] [ frac{d^2z}{dt^2} frac{d}{dt}(3) 0]

Thus, the acceleration vector is:

[ mathbf{a}(t) langle 4, 2, 0 rangle]

Step 4: Evaluating at t 1

Now we can evaluate the velocity and acceleration at :

[ mathbf{v}(1) langle 4(1), 2(1) - 4, 3 rangle langle 4, -2, 3 rangle] [ mathbf{a}(1) langle 4, 2, 0 rangle]

Step 5: Components in a Given Direction

To find the components of the velocity and acceleration vectors in the direction of the vector , we first need to normalize this direction vector:

[ mathbf{d} sqrt{1^2 (-3)^2 2^2} sqrt{1 9 4} sqrt{14}]

The unit direction vector

[ mathbf{u} frac{mathbf{d}}{|mathbf{d}|} leftlangle frac{1}{sqrt{14}}, frac{-3}{sqrt{14}}, frac{2}{sqrt{14}} rightrangle]

Component of Velocity in the Direction of u

The component of the velocity in the direction of

[ mathbf{u} mathbf{v}(1) cdot mathbf{u} langle 4, -2, 3 rangle cdot leftlangle frac{1}{sqrt{14}}, frac{-3}{sqrt{14}}, frac{2}{sqrt{14}} rightrangle] [ frac{1}{sqrt{14}} left( 4 cdot 1 (-2) cdot (-3) 3 cdot 2 right) frac{1}{sqrt{14}} (4 6 6) frac{16}{sqrt{14}}]

Component of Acceleration in the Direction of u

The component of the acceleration in the same direction is calculated in a similar manner:

[ mathbf{a}(1) cdot mathbf{u} langle 4, 2, 0 rangle cdot leftlangle frac{1}{sqrt{14}}, frac{-3}{sqrt{14}}, frac{2}{sqrt{14}} rightrangle] [ frac{1}{sqrt{14}} left( 4 cdot 1 2 cdot (-3) 0 cdot 2 right) frac{1}{sqrt{14}} (4 - 6) frac{-2}{sqrt{14}}]

Conclusion

In this article, we have detailed how to find the components of a particle's velocity and acceleration at a specific time, in a particular direction. This process involves derivatives, vector normalization, and the dot product. Understanding these steps is essential for solving similar problems in vector calculus and related fields.