Question

In Exercises \(25-28,\) a particle is moving along the \(x\) -axis with position function \(x(t) .\) Find the (a) velocity and (b) acceleration, and (c) describe the motion of the particle for \(t \geq 0\) . $$x(t)=t^{2}-4 t+3$$

Short answer

The particle's velocity is given by \(v(t) = 2t - 4\) and its acceleration by \(a(t) = 2\). The particle is moving to the right or positive direction for \(t > 2\) and to the left or negative direction for \(t < 2\). Since the acceleration is always positive, the particle is continuously speeding up for \(t \geq 0\).

Step-by-step solution

Compute Velocity

The velocity of the particle, designated as \(v(t)\), is the first derivative of the position function. So, it is calculated as follows: \(v(t) = x'(t) = 2t - 4\)

Compute Acceleration

The acceleration of the particle, designated as \(a(t)\), is the derivative of the velocity, therefore it is the second derivative of the position function. So, it is calculated as follows: \(a(t) = v'(t) = 2\)

Describe the Motion

The motion of the particle is determined by the signs and values of its velocity and acceleration. Here, the velocity \(v(t) = 2t - 4\) is positive for \(t > 2\) and negative for \(t < 2\), so the particle is moving to the right for \(t > 2\) and to the left for \(t < 2\). The acceleration \(a(t) = 2\) is always positive, suggesting that the particle is speeding up as time increases for \(t \geq 0\)

Key concepts

Velocity and Acceleration

Understanding velocity and acceleration is crucial when studying particle motion in calculus. These two concepts describe how the position of a particle changes over time. Velocity, denoted as v(t), tells us the speed and direction of the particle's movement at a given time; it is the rate of change of the particle's position. Mathematically, we obtain the velocity as the derivative of the position function, \( v(t) = \frac{dx(t)}{dt} \) . For the given function \( x(t) = t^2 - 4t + 3 \), the velocity is \( v(t) = 2t - 4 \).

Acceleration, denoted as a(t), is the rate at which the velocity itself changes. It tells us whether the particle is speeding up or slowing down. This is calculated as the derivative of the velocity function, or the second derivative of the position function, in our case \( a(t) = 2 \). A constant positive acceleration, as seen here, suggests that no matter the initial motion, the particle's speed will consistently increase over time.

For a complete understanding, consider a car in motion: velocity tells you how fast the car is going along the road, while acceleration indicates how the speed of the car is changing - whether it's applying the brakes or stepping on the gas.

Derivative of Position Function

The derivative of the position function provides a powerful tool for analyzing the motion of particles. It's the basis for determining other aspects of motion, such as velocity and acceleration. The position function, usually denoted by \( x(t) \), describes the location of a particle on a coordinate axis at any time t. Taking the derivative of this function with respect to time - \( x'(t) \) - gives us the velocity.

For instance, with the position function \( x(t) = t^2 - 4t + 3 \), the first derivative \( x'(t) = 2t - 4 \) represents the velocity of the particle. The derivative tells us not just how fast the particle is moving but also in which direction - if it's positive, the particle moves to the right on the number line, and if negative, to the left. Taking derivatives allows us to break down complex motion into something more analyzable.

Describing Motion in Calculus

Calculus provides us with a language for describing motion that is both precise and versatile. By using derivatives and the concepts of velocity and acceleration, we can paint a comprehensive picture of how a particle moves. The motion of any object can be understood in terms of its position function, and from there, by analyzing the derivatives, we can determine its velocity and acceleration over time.

Let's take our function \( x(t) = t^2 - 4t + 3 \) as an example. The sign of our velocity equation \( v(t) = 2t - 4 \) changes at t = 2, which indicates a change in direction of the particle's motion. Initially, for t < 2, the velocity is negative, suggesting that the particle is moving to the left. At t = 2, the velocity is zero, implying the particle briefly stops. Beyond that, for t > 2, the particle moves to the right since the velocity is positive.

The constant acceleration of \( a(t) = 2 \) suggests a steady increase in the speed of the particle in the direction of motion. If we were to graph these functions, the motion would become even clearer. This exercise illustrates how calculus can turn an abstract concept of motion into a quantifiable and understandable one.