Question

In Exercises 44 and \(45,\) a particle moves along a line so that its position at any time \(t \geq 0\) is given by \(s(t)=2+7 t-t^{2}\) . Multiple Choice At which of the following times is the particle moving to the left? $$(A)t=0 \quad(\mathbf{B}) t=1 \quad(\mathbf{C}) t=2 \quad(\mathbf{D}) t=7 / 2 \quad(\mathbf{E}) t=4$$

Short answer

The particle is moving to the left at time \(t = 4\) (Option E)

Step-by-step solution

Compute the velocity

The first step is to find the velocity of the particle. The velocity function, \(v(t)\), is the derivative of the position function, \(s(t)\). Therefore, differentiate \(s(t) = 2 + 7t - t^2\) with respect to \(t\) to get the velocity function \(v(t) = 7 - 2t\).

Evaluate the velocity at given times

The next step is to plug in the given time values into the velocity function to check if the particle is moving to the left (velocity is negative). Evaluating the function at each given time: \(v(0) = 7 - 2*0 = 7, \ v(1) = 7 - 2*1 = 5, \ v(2) = 7 - 2*2 = 3, \ v(7/2) = 7 - 2*(7/2) = 0, \ v(4) = 7 - 2*4 = -1 \) . The only negative value is at \(t = 4\).

Conclusion

The only time when the particle is moving to the left (negative velocity) is at \(t = 4\).

Key concepts

Velocity of a Particle

Understanding the velocity of a particle is essential in both the theoretical and practical realms of physics and mathematics. Imagine a particle moving along a straight line where its position at any point in time is represented by a position function, say s(t).

Velocity is the rate at which the particle's position changes over time. Think of it as the speedometer reading in a car, which tells you how fast you're going at a specific moment, but with one additional factor: direction. If the velocity is positive, the particle is moving to the right, while a negative velocity indicates movement to the left.

Mathematically, when you have the position function of an object, such as s(t) = 2 + 7t - t^2, you can obtain the velocity by differentiating this function with respect to time, t. This leads you directly to the concept of the derivative of a position function as a fundamental principle in particle motion calculus.

Derivative of Position Function

To further elucidate the concept of velocity, let's turn our focus onto its mathematical underpinnings, specifically, the derivative of the position function. In calculus, derivatives represent the rate of change of one quantity compared to another.

Every moving object that follows a path has a position that can be graphically plotted and thus can be described with a function with respect to time, s(t). The derivative of this position function is essentially the velocity function v(t). Simply put, it describes how fast the position of the particle changes as time progresses.

In our example, by finding the derivative of s(t) = 2 + 7t - t^2, we obtain the velocity function v(t) = 7 - 2t. Calculating the derivative is essentially peering into the engine of motion, related to how accelerations generate velocities and thus propel the particle along its trajectory. This link between the maths and the physical world exemplifies why calculus is a critical tool in science.

Interpreting Velocity

Once the velocity function is established, interpreting its values is the key to understanding the motion of the particle. In practical terms, interpretation involves assigning meaning to numbers we calculate.

Let's revisit the example where the velocity function v(t) = 7 - 2t has been found. By plugging in different values of time, we can analyze the particle's motion at those specific instances. Remember, a positive value of velocity indicates forward movement, while a negative value signifies the particle is moving backward, or to the left in this one-dimensional context.

At t = 4, our velocity function gives us v(4) = -1. The negative sign divulges that at four seconds, our particle isn't just moving, it's receding to the left. The sign of the velocity is rich with implications: it informs us regarding the direction, and helps in predicting the particle's future course. In the given multiple-choice question, only option E yields negative velocity, making it the clear-cut answer. Interpreting velocity is thus an integral skill for problem-solving in particle motion calculus.