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Chapter 3: Problem 4

give the positions \(s=f(t)\) of a body moving on a coordinate line, with \(s\) in meters and \(t\) in seconds. a. Find the body's displacement and average velocity for the given time interval. b. Find the body's speed and acceleration at the endpoints of the interval. c. When, if ever, during the interval does the body change direction? $$s=\left(t^{4} / 4\right)-t^{3}+t^{2}, \quad 0 \leq t \leq 3$$

Short Answer

Expert verified
a. Displacement: 2.25 m, Average Velocity: 0.75 m/s; b. Speed: 0 m/s at \( t=0 \), 6 m/s at \( t=3 \), Acceleration: 2 m/s虏 at \( t=0 \), 11 m/s虏 at \( t=3 \); c. Direction changes at \( t=1 \) and \( t=2 \).

Step by step solution

01

Understanding Displacement

Displacement is the change in position over the time interval. To find displacement between time \( t = 0 \) and \( t = 3 \), evaluate \( s \) at these points. Calculate \( s(3) \) and \( s(0) \) using the function \( s(t) = \frac{t^4}{4} - t^3 + t^2 \).
02

Calculating Displacement

Evaluate the position function at the endpoints: \[ s(0) = \frac{0^4}{4} - 0^3 + 0^2 = 0 \]\[ s(3) = \frac{3^4}{4} - 3^3 + 3^2 = \frac{81}{4} - 27 + 9 = \frac{9}{4} = 2.25 \]Displacement is \( s(3) - s(0) = 2.25 - 0 = 2.25 \text{ meters}.\)
03

Calculating Average Velocity

Average velocity is the total displacement divided by the time interval. Thus,\[ v_{avg} = \frac{2.25}{3 - 0} = 0.75 \text{ m/s}. \]
04

Understanding Speed and Acceleration at Endpoints

Speed is the magnitude of velocity, and acceleration is the rate of change of velocity. First, find the velocity and acceleration functions by differentiating \( s(t) \).
05

Finding Velocity Function

Differentiate \( s(t) \) to get the velocity function:\[ v(t) = \frac{d}{dt} \left( \frac{t^4}{4} - t^3 + t^2 \right) = t^3 - 3t^2 + 2t. \]
06

Velocity at Endpoints

Evaluate \( v(t) \) at \( t=0 \) and \( t=3 \):\[ v(0) = 0^3 - 3(0)^2 + 2(0) = 0 \]\[ v(3) = 3^3 - 3(3)^2 + 2(3) = 27 - 27 + 6 = 6 \text{ m/s}. \]
07

Finding Acceleration Function

Differentiate \( v(t) \) to get the acceleration function:\[ a(t) = \frac{d}{dt}(t^3 - 3t^2 + 2t) = 3t^2 - 6t + 2. \]
08

Acceleration at Endpoints

Evaluate \( a(t) \) at \( t=0 \) and \( t=3 \):\[ a(0) = 3(0)^2 - 6(0) + 2 = 2 \text{ m/s}^2, \]\[ a(3) = 3(3)^2 - 6(3) + 2 = 27 - 18 + 2 = 11 \text{ m/s}^2. \]
09

Finding When the Body Changes Direction

The body changes direction when its velocity changes sign. Find when \( v(t) = 0 \):Solve for roots in \( v(t) = t^3 - 3t^2 + 2t = t(t^2 - 3t + 2) = 0. \)
10

Points of Velocity Change

Factor \( t^2 - 3t + 2 = (t-1)(t-2) \). The roots are \( t = 0, 1, 2 \). The body changes direction at \( t = 1 \) and \( t = 2 \) since the sign of the velocity changes between these points.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Displacement
Displacement refers to a change in position of an object. It's the overall distance the object moves from its initial position to its final position along a straight line. In calculus, displacement can be determined using a function that describes the position of the object as it moves over time. By evaluating this function at different points in time, we can calculate the displacement.
  • To find the displacement, you look at the position function evaluated at the start and end times.
  • For the function given: \[ s(t) = \frac{t^4}{4} - t^3 + t^2 \], the displacement between times \( t = 0 \) and \( t = 3 \) is simply the difference \( s(3) - s(0) \).
  • After evaluating, we find \( s(3) = 2.25 \) meters and \( s(0) = 0 \) meters, leading to a displacement of \( 2.25 \) meters.
Average Velocity
Average velocity indicates how fast an object is moving on average over a given time interval. It represents the ratio of displacement to the total time taken. This is different from speed, which does not consider direction.
  • Average velocity is calculated by dividing the displacement by the time interval during which the displacement occurs.
  • For the problem at hand, with a displacement of \( 2.25 \) meters over \( 3 \) seconds, the average velocity is \( v_{avg} = \frac{2.25}{3} = 0.75 \) m/s.
  • It's important to note that average velocity is a vector quantity, meaning it has both magnitude and direction.
Acceleration
Acceleration is crucial to understanding how the speed of an object changes over time. Simply put, it's the rate at which velocity changes. Calculus helps us find acceleration by differentiating the velocity function.
  • To find acceleration, differentiate the velocity function \( v(t) \) which is \( t^3 - 3t^2 + 2t \).
  • This results in the acceleration function \( a(t) = 3t^2 - 6t + 2 \).
  • To understand how fast the speed is changing at any specific time, evaluate \( a(t) \) at that point. At times \( t=0 \) and \( t=3 \), we assess: \( a(0) = 2 \text{ m/s}^2 \) and \( a(3) = 11 \text{ m/s}^2 \).
  • This shows how the object's acceleration changes from the starting point of the interval to the endpoint.

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