Question

A particle moving along a straight line has a velocity of \(v(t)=t^{2} e^{-t}\) after \(t \mathrm{sec}\). How far does it travel in the first 2 sec? (Assume the units are in feet and express the answer in exact form.)

Short answer

The particle travels \(2e^{-2} - 2\) feet in the first 2 seconds.

Step-by-step solution

Understand the Problem

We need to determine the distance traveled by the particle over the first 2 seconds, given its velocity function \(v(t) = t^2 e^{-t}\). The distance traveled is the definite integral of the velocity function from 0 to 2.

Set up the Integral

To find the distance traveled, we need to compute the integral of the velocity function from 0 to 2:\[\int_0^2 t^2 e^{-t} \, dt\]

Choose an Integration Technique

The integrand \(t^2 e^{-t}\) suggests using integration by parts. Integration by parts is derived from the product rule for differentiation and is given by:\[\int u \, dv = uv - \int v \, du\]We will choose \(u = t^2\) and \(dv = e^{-t} \, dt\).

Apply Integration by Parts

First, find \(du\) and \(v\):- \(u = t^2\), so \(du = 2t \, dt\).- \(dv = e^{-t} \, dt\), so \(v = -e^{-t}\).Applying integration by parts:\[\int t^2 e^{-t} \, dt = -t^2 e^{-t} - \int -2t e^{-t} \, dt\]

Repeat Integration by Parts

We now have another integral that also requires integration by parts:\[\int 2t e^{-t} \, dt\]Set \(u_1 = 2t\) and \(dv_1 = e^{-t} \, dt\).- \(du_1 = 2 \, dt\)- \(v_1 = -e^{-t}\)Applying integration by parts again:\[-2 \int t e^{-t} \, dt = -2(-t e^{-t} + \int e^{-t} \, dt) = 2t e^{-t} - 2e^{-t}\]

Compute the Integral

Combine the terms from both integration by parts applications:\[-t^2 e^{-t} + 2t e^{-t} - 2e^{-t} = (-t^2 + 2t - 2) e^{-t}\]Now evaluate this from 0 to 2:

Evaluate the Definite Integral

Evaluate at the bounds:\[((-2)^2 - 2 \times 2 + 2) e^{-2} - (0^2 - 0 \times 2 + 2) e^{0}= (4 - 4 + 2) e^{-2} - 2 = 2e^{-2} - 2\]Thus, the integral evaluates to \(2e^{-2} - 2\).

Express the Answer in Exact Form

The exact distance traveled by the particle in the first 2 seconds is \[2e^{-2} - 2\] feet.

Key concepts

Understanding the concept of a definite integral

In calculus, a definite integral is a way to calculate the area under a curve using integration. When dealing with a velocity function, like in the given problem, calculating the definite integral helps in determining the total distance traveled by a particle over a certain period of time.

The definite integral is represented as \[ \int_a^b f(x) \, dx \]where \( a \) and \( b \) are the bounds. This integral calculates the accumulation of the quantity \( f(x) \) from \( a \) to \( b \).

• The starting point \( a \) corresponds to the initial time.
• The endpoint \( b \) corresponds to the final time.
• \( f(x) \) represents the function being integrated—such as velocity in motion problems.

In this exercise, the expression \( \int_0^2 t^2 e^{-t} \, dt \) is evaluated to find the total distance the particle travels from \( t = 0 \) to \( t = 2 \) seconds.

Applying integration by parts

Integration by parts is an indispensable technique in calculus, especially for integrals involving a product of functions. It is built on the product rule for differentiation and used to simplify complex integrals. If you have a product of two functions, you can split these functions into parts and use integration by parts.

The formula for integration by parts is given as:\[ \int u \, dv = uv - \int v \, du \]where:

• \( u \) is a function you choose to differentiate,
• \( dv \) is a part of the original function you integrate,
• \( du \) is the derivative of \( u \), and
• \( v \) is the integral of \( dv \).

In solving the integral \( \int t^2 e^{-t} \, dt \), choosing \( u = t^2 \) and \( dv = e^{-t} \, dt \) simplifies the process. It involves differentiation of the polynomial term and integration of the exponential term—ideal for reducing the complexity of evaluation. This method yields new integrals that are easier to solve.

Exploring the velocity function

A velocity function, \( v(t) \), describes how the speed of an object changes over time. It determines both the rate of motion and the direction. In calculus problems involving motion, it plays a critical role in identifying how far an object has traveled over a given time period.

In the given exercise, the velocity function is \( v(t) = t^2 e^{-t} \). Let's break it down:

• The part \( t^2 \) suggests that the velocity changes quadratically with time.
• The \( e^{-t} \) term implies that velocity decreases exponentially as time progresses.

The combination of polynomial \( t^2 \) and exponential \( e^{-t} \) features in the velocity function explains the necessity of using integration by parts, given their mixed behavior in calculus. Understanding how the function behaves helps calculate the integral, providing the distance the particle travels during the timeframe specified—crucial in physics to grasp real-world motion dynamics.