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Magazine Chapter 5: Problem 12
A particle moves with a velocity of \(v(t) \mathrm{m} / \mathrm{s}\) along an \(\mathrm{s}\) -axis. Find the displacement and the distance traveled by the particle during the given time interval. $$ \begin{array}{l}{\text { (a) } v(t)=t-\sqrt{t} ; 0 \leq t \leq 4} \\ {\text { (b) } v(t)=\frac{1}{\sqrt{t+1}} ; 0 \leq t \leq 3}\end{array} $$
Short Answer
Expert verified
(a) Displacement: \(\frac{8}{3}\) m; Distance: 4 m.
(b) Displacement and Distance: 2 m.
Step by step solution
01
Understanding the Problem
We need to find the displacement and distance traveled by a particle with a given velocity function during a specified time interval. Displacement can be found by integrating the velocity function over the interval, while distance is calculated by integrating the absolute value of the velocity function over the same interval.
02
Part (a) - Express Displacement
For part (a), the velocity function is given by \(v(t) = t - \sqrt{t}\). First, we'll compute the displacement, which is the integral of \(v(t)\) from \(t = 0\) to \(t = 4\):\[\text{Displacement} = \int_{0}^{4} (t - \sqrt{t}) \, dt\].
03
Part (a) - Compute the Integral
Evaluate the integral step-by-step:1. Integrate \(t\): \(\int t \, dt = \frac{t^2}{2}\).2. Integrate \(\sqrt{t}\): \(\int \sqrt{t} \, dt = \frac{2}{3} t^{3/2}\).Combine these expressions:\[\int_0^4 (t - \sqrt{t}) \, dt = \left[ \frac{t^2}{2} - \frac{2}{3}t^{3/2} \right]_{0}^{4}\].
04
Part (a) - Evaluate Definite Integral
Calculate the definite integral:\[\left(\frac{4^2}{2} - \frac{2}{3} (4)^{3/2}\right) - \left(\frac{0^2}{2} - \frac{2}{3} (0)^{3/2}\right)\]Simplify:\[= \left(8 - \frac{2}{3} \cdot 8\right) - 0 = 8 - \frac{16}{3} = \frac{8}{3}\].The displacement is \(\frac{8}{3}\) meters.
05
Part (a) - Calculate Distance
To find the distance, integrate \(|v(t)|\). Note that \(v(t)\) changes sign within \([0, 4]\) at \(t = 1\). Thus, calculate:- \(\int_{0}^{1} |t - \sqrt{t}| \, dt\) and \(\int_{1}^{4} |t - \sqrt{t}| \, dt\).- Since \(v(t) < 0\) on \([0, 1]\) and \(v(t) > 0\) on \([1, 4]\), we have:\[\int_{0}^{1} (\sqrt{t} - t) \, dt + \int_{1}^{4} (t - \sqrt{t}) \, dt\].
06
Part (a) - Evaluate Distance Integrals
Evaluate both integrals:1. \(\int_{0}^{1} (\sqrt{t} - t) \, dt = \left[ \frac{2}{3}t^{3/2} - \frac{t^2}{2} \right]_{0}^{1} = \frac{2}{3} - 0.5 = \frac{1}{6}\).2. The second integral is the one already computed for displacement on \([1, 4]\): \(\frac{23}{6} - \frac{2}{3}\).Add them:\[\frac{1}{6} + \frac{23}{6} = 4\].Thus, the distance is 4 meters.
07
Part (b) - Express Displacement
Now, consider part (b) where the velocity function is \(v(t) = \frac{1}{\sqrt{t+1}}\). The displacement is:\[\text{Displacement} = \int_{0}^{3} \frac{1}{\sqrt{t+1}} \, dt\].
08
Part (b) - Compute the Integral
The integral \(\int \frac{1}{\sqrt{t+1}} \, dt\) simplifies to solving \(u = t+1\) with \(du = dt\). Integrating gives:\[2\sqrt{t+1}\] over \([0, 3]\). Thus, evaluate:\[= 2 \sqrt{4} - 2\sqrt{1} = 4 - 2 = 2\].
09
Part (b) - Calculate Distance
Since \(v(t) = \frac{1}{\sqrt{t+1}}\) is non-negative for all \(t \geq 0\), the distance traveled is the same as the displacement. Hence:\[\text{Distance} = \int_{0}^{3} \frac{1}{\sqrt{t+1}} \, dt = 2\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Displacement in Calculus
Displacement in calculus refers to the amount of change in position of a particle. It doesn't matter what path it takes; only the initial and final positions are important. We use calculus to find displacement by integrating the velocity function over a set time period. For example, if a particle moves along a straight line with the velocity function \(v(t)\), the displacement over the time interval \(a, b\) is determined by finding the definite integral:\[ \text{Displacement} = \int_{a}^{b} v(t) \, dt \] This process involves finding the antiderivative of the velocity function first, then evaluating it at the endpoints of the interval. The displacement can be positive, negative, or zero, indicating the direction of motion relative to the starting point.
Integrating Velocity Functions
Integrating velocity functions is a fundamental skill in calculus used to determine displacement. To integrate a velocity function, we perform definite integration over the given interval.
Velocity functions can be composed of simple terms, such as polynomials or more complex ones involving roots or fractions. Let's break down the steps involved in integrating:
- Compute the integral: Find the antiderivative of the velocity function.
- Evaluate the integral: Substitute the limits of integration to compute the net area under the velocity-time graph.
Calculating Distance Traveled
Calculating the distance traveled by a particle involves capturing the total ground covered, regardless of direction. Unlike displacement, distance takes all movement into account. To compute the distance traveled when given a velocity function that changes sign, we use:\[\text{Distance} = \int_{a}^{b} |v(t)| \, dt\] This involves finding the integral over the interval but using the absolute value of the velocity function. Steps to calculate the distance traveled:
- Identify where the velocity changes sign in the interval.
- Split the integral at those points.
- Integrate each interval separately, taking the absolute value of the velocity function to ensure all areas add positively.
