Problem

The linear density of a rod of length $ 4 m $ is …

View

Problem 62 Hard Difficulty

The acceleration function (in $ m/s^2 $) and the initial velocity are given for a particle moving along a line. Find (a) the velocity at time $ t $ and (b) the distance traveled during the given time interval.

$ a(t) = 2t + 3 $, $ v(0) = -4 $, $ 0 \le t \le 3 $

Answer

a) $t^{2}+3 t-4$
b) $\frac{-1}{3}-\frac{3}{2}+4 +9+\frac{21}{2}-12 -\frac{1}{3}-\frac{3}{2}+4$

Watch More Solved Questions in Chapter 5

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Video Transcript

Alright I think give us the acceleration is to T plus three to t plus three then. Uh And we also know that the derivative of velocity is equal to acceleration. So if I find the integral of that uh it will give me the velocity so that now I. M. V. Of T. And the anti directive we add one to the expo divided by your new X 12 by by two is one Plus three T Plus C. And that's where we're going to use B. Uh Zero is equal to negative four because as I plug in negative four and for velocity and plug in zero for these two T 00 squared plus three times zero is just zero. Your left was C. Is equal to negative four. So the equation of velocity would be T squared plus three t minus four. Now as far as total distance traveled goes um Probably what I would do is first of all we care about from 0 to 3 zero 23 Um as I would see if there's any change in direction and you find the changing direction by setting this equal to zero. If you find factors in mega for that had to be three before -1. So there will be T plus four t minus one. So this object changes directions at T equals negative for which doesn't really make any sense and T.E. is one. Um Let's see. So what would I do? I would you're gonna have to find the integral of this, which is why this problem is so complicated to find that's right T D. T. The total distance traveled. That integral from 0 to 1 of that function. GT Yeah. Starting here and then it changes directions. Uh and then the integral from 1 to 3 of t squared plus three t minus four. Yeah. Gt uh Now one of these answers to be negative because we're moving to the left I don't know which one that is. I'm just doing the anti derivative From 0 to wine and then the same thing here really all the same work but as you plug in one you'll get one third plus three halves minus four. Yeah that's going to be a negative number. So absolute value that thing because total additions traveler needs to be positive and then add to it. This I guess I didn't show plugging in zero but you get a bunch of zeros. Um One third plus three halves one third times three squared cubed times three halves times three squared minus four times three. You can see why there's so much work to be done here minus plugging in all those ones now. On third plus three hands minus four. Yeah. Yeah. Uh I'll go ahead and put absolute value there just in case. Um And let's see yeah we can go to calculate and just type that in. Mhm. See absolute value that and then we squared Plus three cubed divided by two minus slopes. I'm using a calculator and I'm struggling right now minus 12 minus the quantity. One third plus 3/2 -4 hopefully attacked us all incorrectly because I don't want to go back and change anything. Um And I can change that to a decimal or to a fraction as well. Math track will say 89 6. Here we go.

Gregory H.

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Problem

The linear density of a rod of length $ 4 m $ is …

Problem 62 Hard Difficulty

The acceleration function (in $ m/s^2 $) and the initial velocity are given for a particle moving along a line. Find (a) the velocity at time $ t $ and (b) the distance traveled during the given time interval.

$ a(t) = 2t + 3 $, $ v(0) = -4 $, $ 0 \le t \le 3 $

Answer

a) $t^{2}+3 t-4$
b) $\frac{-1}{3}-\frac{3}{2}+4 +9+\frac{21}{2}-12 -\frac{1}{3}-\frac{3}{2}+4$

More Answers

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 1 / AB Educators

Catherine R.

Caleb E.

Samuel H.

Michael J.

Calculus 1 / AB Courses

Precalculus Review - Intro

Functions on the Real Line - Overview

Recommended Videos

Watch More Solved Questions in Chapter 5

Video Transcript

Calculus: Early Transcendentals

Catherine R.

Caleb E.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

a) $t^{2}+3 t-4$b) $\frac{-1}{3}-\frac{3}{2}+4 +9+\frac{21}{2}-12 -\frac{1}{3}-\frac{3}{2}+4$

Calculus: Early Transcendentals

Catherine R.

Caleb E.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

Catherine R.

Caleb E.

Samuel H.

Michael J.

a) $t^{2}+3 t-4$
b) $\frac{-1}{3}-\frac{3}{2}+4 +9+\frac{21}{2}-12 -\frac{1}{3}-\frac{3}{2}+4$