Question

The velocity of a particle moving along a line is a function of time given by \(v(t)=\frac{88 t^{2}}{t^{2}+1} .\) Find the distance that the particle has traveled after \(t=5 \mathrm{sec}\).

Short answer

The particle traveled approximately 319.14 units after 5 seconds.

Step-by-step solution

Understand the Problem

We're given a velocity function of a particle, \(v(t)=\frac{88 t^{2}}{t^{2}+1}\), and we need to find the distance the particle has traveled after \(t = 5\) seconds. The distance traveled can be found by integrating the velocity function over the given time interval.

Set up the Integral

To find the distance traveled, we need to calculate the integral of the velocity function from \(t = 0\) to \(t = 5\). This is because distance is the integral of velocity over time: \[ \text{Distance} = \int_0^5 v(t) \, dt = \int_0^5 \frac{88 t^2}{t^2 + 1} \, dt\]

Solve the Integral

We need to evaluate the integral \(\int_0^5 \frac{88 t^2}{t^2 + 1} \, dt\). This can be done by simplifying or using partial fraction decomposition, but here it's straightforward. Notice that the integral \(\int \frac{t^2}{t^2 + 1} \, dt\) can be solved using substitution or known antiderivatives. However, in this case, an immediate antiderivative approach leads to:\[88\int_0^5 \left(1 - \frac{1}{t^2 + 1}\right) \, dt = 88\left(\left[t - \tan^{-1}(t)\right]_0^5\right)\]Apply limits to find distance.

Evaluate the Integral and Calculate Distance

Now substitute the limits into the evaluated expression: \[88\left( \left[5 - \tan^{-1}(5)\right] - \left[0 - \tan^{-1}(0)\right] \right)\]This simplifies to:\[88 \left(5 - \tan^{-1}(5) + 0\right)\]Using \(\tan^{-1}(0) = 0\), we find:\[88(5 - \text{tan}^{-1}(5))\] Using calculators or tables, approximate \(\tan^{-1}(5) \approx 1.3734\).\[88 \cdot (5 - 1.3734) \approx 88 \times 3.6266\]Calculate this result.

Find the Final Result

Perform the final calculation:\[88 \times 3.6266 \approx 319.1408\]Thus, the distance traveled by the particle after \(5\) seconds is approximately \(319.14\) units.

Key concepts

Velocity Function

In integral calculus, velocity functions are essential for understanding how fast an object moves at any given time. The velocity function for a particle can vary with time, as seen in our example: \[ v(t) = \frac{88 t^{2}}{t^{2}+1} \] This function provides the velocity at any time \( t \). By plotting this function, one could visualize how the velocity changes. It's crucial to understand that the velocity function represents the rate of change of the particle's position over time. To comprehend what this means:

• A higher function value indicates faster movement.
• A zero value suggests a moment of rest.

In our case, as time progresses, the function's behavior governs the speed of the particle along its path. This concept forms the base when using calculus to connect motion and change.

Distance Calculation

Distance calculation in integral calculus involves integrating the velocity function. This method accumulates all the little pieces of travel over a time span. If you imagine time divided into tiny intervals, each interval's contribution adds up to the total distance. In our example, this is represented by the integral: \[ \text{Distance} = \int_0^5 v(t) \, dt = \int_0^5 \frac{88 t^2}{t^2 + 1} \, dt \] Why integrate?

• Integration aggregates the instantaneous velocities over time.
• The entire shaded area under the curve of \( v(t) \) on a graph represents total distance.

Thus, evaluating this integral over the given limits helps us determine how far the particle has moved in the specified time.

Antiderivatives

The concept of antiderivatives is a cornerstone in integral calculus. An antiderivative of a function is another function whose derivative gives back the original function. For velocity functions, finding the antiderivative is a way to obtain a displacement function. In solving the integral of our velocity function: \[ \int \frac{88 t^2}{t^2 + 1} \, dt \] We look for a function whose derivative brings us back to the given velocity function. By utilizing known calculus techniques or substitutions, an immediate antiderivative approach gave us: \[ 88\left(t - \tan^{-1}(t)\right) \] This function allows us to substitute in limits (from 0 to 5 in our case) and find the relation of change. Thus, antiderivatives serve as a bridge from velocity to other physical quantities like displacement or distance.

Definite Integral

A definite integral provides an actual numerical value that represents the accumulated total change over a specified interval. This is a practical application in calculating distances, areas, and other physical characteristics. For calculating the particle's traveled distance, the definite integral approach provides: \[ \int_0^5 \frac{88 t^2}{t^2 + 1} \, dt = 88\left(5 - \tan^{-1}(5) + 0\right) \] Using the limits from 0 to 5, we plugged them into our antiderivative expression. This leads to a final value: \[ 88 \times 3.6266 \approx 319.1408 \] The definite integral results in the numerical distance of approximately 319.14 units. Thus, definite integrals give us tangible values, reflecting accumulated effects over designated spans.