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Explain and discuss the difference in velocity at times \(t=1\) and \(t=5\) for \(v(t)=2 t-t^{2}\).

Short Answer

Expert verified
At \(t=1\), velocity is 1 unit; at \(t=5\), velocity is -15 units, showing a direction reversal and increased speed.

Step by step solution

01

Define the Velocity Function

The velocity of the object is given by the function \(v(t) = 2t - t^2\). This formula represents the velocity at any time \(t\). We will use this to find the velocities at specific times.
02

Compute Velocity at Time t=1

Substitute \(t = 1\) into the velocity function to find the velocity at this moment.\[v(1) = 2(1) - (1)^2 = 2 - 1 = 1.\]So, the velocity at \(t = 1\) is 1 unit.
03

Compute Velocity at Time t=5

Substitute \(t = 5\) into the velocity function to find the velocity at \(t = 5\).\[v(5) = 2(5) - (5)^2 = 10 - 25 = -15.\]Thus, the velocity at \(t = 5\) is -15 units.
04

Analyze the Velocity Change

At \(t=1\), the object has a positive velocity of 1 unit, indicating it's moving in the positive direction. By \(t=5\), the velocity is -15 units, indicating the object has reversed direction and is moving negatively at a faster speed.

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Key Concepts

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

Understanding the Velocity Function
In calculus, velocity functions help us determine how an object moves over time. A velocity function describes how fast and in which direction an object is moving at any given time. The velocity function in our original exercise is given by:
  • \( v(t) = 2t - t^2 \)
In this function, the variable \( t \) represents time. The function itself calculates velocity based on the time input you provide. It's important to note that the terms in the velocity function might involve operations like multiplication or squaring, which can significantly affect the outcome depending on the value of \( t \). For instance, the term \(-t^2\) decreases the velocity as \( t \) increases because it becomes more negative.
Understanding the components of a velocity function is essential in predicting and analyzing the movement of objects. This function can tell you whether an object is accelerating, decelerating, moving forwards, or even backwards.
Exploring Velocity Change
Velocity change occurs when there is a difference in velocity at different times. In the original exercise, we witnessed a change in velocity between the times \( t = 1 \) and \( t = 5 \). At \( t = 1 \), we determined the velocity to be 1 unit, and at \( t = 5 \), it dropped to -15 units.

This change signifies that the object not only slowed down but also reversed its direction. Here is what happened:
  • At \( t = 1 \), the object moves positively with a speed of 1 unit.
  • By \( t = 5 \), speed and direction have changed entirely, moving negatively with a speed of -15 units.
Understanding these changes is crucial when studying motion, as it explains how an object's speed and trajectory can alter over time due to external and internal factors. Noticing such patterns of velocity change can further help in forecasting future positions or conditions of the object.
Evaluating Functions at Specific Points
Evaluating functions at specific points is a fundamental skill in mathematics, making it possible to find exact values of functions. This process involves replacing the variable in a function with a specific number. In our case, we plugged specific times into the velocity function to understand movement.
  • For \( t = 1 \), we substituted the time into the equation \( v(t) = 2(1) - (1)^2 = 1 \).
  • For \( t = 5 \), we followed with \( v(t) = 2(5) - (5)^2 = -15 \).
Calculating these values gives us snapshots of the object's condition at those times. This concept doesn't solely apply to velocity functions but to any function where a change in the independent variable may lead to different outputs. Evaluating functions this way helps with understanding broader mathematical and physical relationships, such as acceleration, displacement, and beyond. So, while this technique may seem simple, it serves as a powerful tool across various fields of study.

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Most popular questions from this chapter

A ball is thrown upward, and its height, \(h,\) in metres above the ground after \(t\) seconds is given by \(h(t)=-5 t^{2}+25 t, t \geq 0\) a. Calculate the ball's initial velocity. b. Calculate its maximum height. c. When does the ball strike the ground, and what is its velocity at this time?

The polynomial function \(f(x)=0.001 x^{3}-0.12 x^{2}+3.6 x+10,0 \leq x \leq 75\) models the shape of a roller-coaster track, where \(f\) is the vertical displacement of the track and \(x\) is the horizontal displacement of the track. Both displacements are in metres. Determine the absolute maximum and minimum heights along this stretch of track.

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A north-south highway intersects an east-west highway at point \(P .\) A vehicle crosses \(P\) at \(1: 00 \mathrm{p} . \mathrm{m} .,\) travelling east at a constant speed of \(60 \mathrm{km} / \mathrm{h} .\) At the same instant, another vehicle is \(5 \mathrm{km}\) north of \(P,\) travelling south at \(80 \mathrm{km} / \mathrm{h}\). Find the time when the two vehicles are closest to each other and the distance between them at this time.

Starting at \(t=0,\) a particle moves along a line so that its position after \(t\) seconds is \(s(t)=t^{2}-6 t+8,\) where \(s\) is in metres. a. What is its velocity at time \(t ?\) b. When is its velocity zero?

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