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Magazine Chapter 5: Problem 17
In each part, use the given information to find the position, velocity, speed, and acceleration at time \(t=1\). (a) \(v=\sin \frac{1}{2} \pi t ; s=0\) when \(t=0\) (b) \(a=-3 t ; s=1\) and \(v=0\) when \(t=0\)
Short Answer
Expert verified
(a) Position: 0, Velocity: 1, Speed: 1, Acceleration: 0.
(b) Position: 0.5, Velocity: -1.5, Speed: 1.5, Acceleration: -3.
Step by step solution
01
Problem Analysis for Part (a)
We are given the velocity function \(v(t) = \sin \frac{1}{2} \pi t\) and the initial position \(s(0) = 0\). Our goals are to find the position, speed, and acceleration at \(t = 1\).
02
Finding Velocity at t=1 for Part (a)
First, plug \(t = 1\) into the velocity function. Thus, \(v(1) = \sin \frac{1}{2} \pi \times 1 = \sin \frac{\pi}{2}\). Knowing that \(\sin \frac{\pi}{2} = 1\), we have \(v(1) = 1\).
03
Calculating Position at t=1 for Part (a)
To find position, integrate the velocity function: \(s(t) = \int \sin \frac{1}{2} \pi t \, dt\). This gives \(s(t) = -\frac{2}{\pi} \cos \frac{1}{2} \pi t + C\). Using the initial condition \(s(0) = 0\), solve for \(C\): \[-\frac{2}{\pi} \cos(0) + C = 0 \Rightarrow C = \frac{2}{\pi}\].The position function is \(s(t) = -\frac{2}{\pi} \cos \frac{1}{2} \pi t + \frac{2}{\pi}\). Evaluate at \(t = 1\): \(s(1) = 0\).
04
Determining Speed for Part (a)
Speed is the absolute value of velocity. Since \(v(1) = 1\), the speed at \(t = 1\) is also \(1\).
05
Finding Acceleration at t=1 for Part (a)
Acceleration is the derivative of velocity. So, differentiate \(v(t) = \sin \frac{1}{2} \pi t\) to get \(a(t) = \frac{1}{2} \pi \cos \frac{1}{2} \pi t\). Evaluate at \(t = 1\): \(a(1) = \frac{1}{2} \pi \times \cos \frac{\pi}{2} = 0\).
06
Problem Analysis for Part (b)
For this part, we know acceleration \(a(t) = -3t\) with initial conditions \(s(0) = 1\) and \(v(0) = 0\). Our task is to find the position, velocity, speed, and acceleration at \(t = 1\).
07
Calculating Velocity Function for Part (b)
Integrate acceleration to get velocity: \(v(t) = \int -3t \, dt = -\frac{3}{2}t^2 + C\). Use the condition \(v(0) = 0\) to find \(C\): \(-\frac{3}{2} \times 0 + C = 0 \Rightarrow C = 0\). Thus, \(v(t) = -\frac{3}{2} t^2\). Evaluate at \(t = 1\): \(v(1) = -\frac{3}{2} \times 1^2 = -\frac{3}{2}\).
08
Determining Speed for Part (b)
Speed is the absolute value of velocity. Since \(v(1) = -\frac{3}{2}\), speed is \(\frac{3}{2}\).
09
Calculating Position Function for Part (b)
Integrate velocity to get position: \(s(t) = \int -\frac{3}{2} t^2 \, dt = -\frac{1}{2} t^3 + D\).With \(s(0) = 1\), solve for \(D\): \(-\frac{1}{2} \times 0 + D = 1 \Rightarrow D = 1\).Therefore, \(s(t) = -\frac{1}{2} t^3 + 1\). Evaluate at \(t = 1\): \(s(1) = -\frac{1}{2} \times 1^3 + 1 = \frac{1}{2}\).
10
Finding Acceleration at t=1 for Part (b)
Acceleration is given as \(a(t) = -3t\). Evaluating at \(t = 1\) gives \(a(1) = -3 \times 1 = -3\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Velocity
Velocity is the rate at which an object changes its position. It is a vector quantity, meaning it has both magnitude and direction. In calculus, the velocity function is often derived from the rate of change of position with respect to time.
- Mathematically, velocity is expressed as the derivative of the position function, denoted as \(v(t) = \frac{ds}{dt}\), where \(s\) represents position and \(t\) represents time.
- It can also be viewed in terms of a function of time, such as in part (a) of our exercise where \(v(t) = \sin \frac{1}{2} \pi t\).
- The velocity at a specific time, such as \(t = 1\), can be found by substituting the time into the velocity function.
Position
Position indicates the location of a particle or an object along a path in a particular time instant. It's an essential concept that gets attributed to an object's placement along a coordinate system.
- The position function \(s(t)\) describes where an object is at any given time \(t\).
- To determine this, you often need to integrate a given velocity function, reflecting that position is the cumulative sum of velocities over time.
- For example, in the exercise, finding the position involved integrating the velocity \(v(t)\), resulting in the function \(s(t) = -\frac{2}{\pi} \cos \frac{1}{2} \pi t + \frac{2}{\pi}\).
Acceleration
Acceleration is the rate of change of velocity with respect to time. Like velocity, it is a vector, possessing both magnitude and direction.
- In calculus, acceleration is determined as the derivative of the velocity function, represented as \(a(t) = \frac{dv}{dt}\).
- In our problems, it was demonstrated through examples where the acceleration function provides insight into how quickly an object speeds up or slows down. For instance, in part (b), \(a(t) = -3t\) indicates constantly changing acceleration.
- Evaluating acceleration at a particular time, such as \(t = 1\), gives an immediate snapshot of the rate of velocity change at that specific point.
