s = f(t) = t³ - 12t² + 36t ds v(t) = 3²- - 24t+ 36 dt = The velocity after 2 s means the instantaneous velocity when t = 2, that is ds v(2)= dt = 2 The velocity after 3 s is = 3(2)²-24(2) + 36 = m/s.
s = f(t) = t³ - 12t² + 36t ds v(t) = 3²- - 24t+ 36 dt = The velocity after 2 s means the instantaneous velocity when t = 2, that is ds v(2)= dt = 2 The velocity after 3 s is = 3(2)²-24(2) + 36 = m/s.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Fast pls solve this question correctly in 5 min pls I will give u like for sure
Sini
![(a) The velocity function is the derivative of the position function.
s = f(t) = t³ - 12t² + 36t
ds
dt
v(t) = - 3t² - 24t+ 36
=
=
(b) The velocity after 2 s means the instantaneous velocity when t = 2, that is
v(2):
= $12-2²
ds
dt
t=2
The velocity after 3 s is
v(3) = 3(3)² - 24(3) + 36 =
t=6
s=0
that is,
(c) The particle is at rest when v(t)
3t² - 24t+ 36 = 3(t² − 8t + 12) = 3(t − 2)(t - 6) =
(smaller t-value) or t =
and this is true when t =
and after
t=0
S=0
=
= 3(2)2 24(2) + 36 =
|f(2)=f(0)| =
|f(6) - f(2)=
=
(d) The particle moves in the positive direction when v(t) ?
3t² - 24t+ 36 = 3(t-2)(t-6) ? ⁰.
=
This inequality is true when both factors are positive (t >
the positive direction in the time intervals t < t₁
t₁ = [
<t<
|= t₂.
(e) Using the information from part (d) we make a schematic sketch in the figure of the motion of the particle back and forth along a line (the s-axis).
=
m/s.
s (larger t-value).
=
t=2
s=32
From t = 2 to t = 6 the distance traveled is
m/s.
=
m.
m.
From t = 6 to t = 8 the distance traveled is
|f(8) - f(6)|
The total distance traveled in the first eight seconds is
(f) Because of what we learned in parts (d) and (e), we need to calculate the distances traveled during the time intervals [0, 2], [2, 6], [6, 8]
separately.
The distance traveled in the first 2 seconds is
(larger t-value). Thus, the particle is at rest after
m.
0, that is,
) or when both factors are negative (t <
and t >
m.
t₂
s (smaller t-value)
=
). Thus the particle moves in
It moves backward (in the negative direction) when](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd8f73ea3-8b9d-41a5-8140-775c62a3c253%2Fa3bfdec3-5066-4f33-998b-a0ded6231014%2Fdh0ux4_processed.png&w=3840&q=75)
Transcribed Image Text:(a) The velocity function is the derivative of the position function.
s = f(t) = t³ - 12t² + 36t
ds
dt
v(t) = - 3t² - 24t+ 36
=
=
(b) The velocity after 2 s means the instantaneous velocity when t = 2, that is
v(2):
= $12-2²
ds
dt
t=2
The velocity after 3 s is
v(3) = 3(3)² - 24(3) + 36 =
t=6
s=0
that is,
(c) The particle is at rest when v(t)
3t² - 24t+ 36 = 3(t² − 8t + 12) = 3(t − 2)(t - 6) =
(smaller t-value) or t =
and this is true when t =
and after
t=0
S=0
=
= 3(2)2 24(2) + 36 =
|f(2)=f(0)| =
|f(6) - f(2)=
=
(d) The particle moves in the positive direction when v(t) ?
3t² - 24t+ 36 = 3(t-2)(t-6) ? ⁰.
=
This inequality is true when both factors are positive (t >
the positive direction in the time intervals t < t₁
t₁ = [
<t<
|= t₂.
(e) Using the information from part (d) we make a schematic sketch in the figure of the motion of the particle back and forth along a line (the s-axis).
=
m/s.
s (larger t-value).
=
t=2
s=32
From t = 2 to t = 6 the distance traveled is
m/s.
=
m.
m.
From t = 6 to t = 8 the distance traveled is
|f(8) - f(6)|
The total distance traveled in the first eight seconds is
(f) Because of what we learned in parts (d) and (e), we need to calculate the distances traveled during the time intervals [0, 2], [2, 6], [6, 8]
separately.
The distance traveled in the first 2 seconds is
(larger t-value). Thus, the particle is at rest after
m.
0, that is,
) or when both factors are negative (t <
and t >
m.
t₂
s (smaller t-value)
=
). Thus the particle moves in
It moves backward (in the negative direction) when
![(a) The velocity function is the derivative of the position function.
s = f(t) = t³ - 12t² + 36t
ds
dt
v(t) = - 3t² - 24t+ 36
=
=
(b) The velocity after 2 s means the instantaneous velocity when t = 2, that is
v(2):
= $12-2²
ds
dt
t=2
The velocity after 3 s is
v(3) = 3(3)² - 24(3) + 36 =
t=6
s=0
that is,
(c) The particle is at rest when v(t)
3t² - 24t+ 36 = 3(t² − 8t + 12) = 3(t − 2)(t - 6) =
(smaller t-value) or t =
and this is true when t =
and after
t=0
S=0
=
= 3(2)2 24(2) + 36 =
|f(2)=f(0)| =
|f(6) - f(2)=
=
(d) The particle moves in the positive direction when v(t) ?
3t² - 24t+ 36 = 3(t-2)(t-6) ? ⁰.
=
This inequality is true when both factors are positive (t >
the positive direction in the time intervals t < t₁
t₁ = [
<t<
|= t₂.
(e) Using the information from part (d) we make a schematic sketch in the figure of the motion of the particle back and forth along a line (the s-axis).
=
m/s.
s (larger t-value).
=
t=2
s=32
From t = 2 to t = 6 the distance traveled is
m/s.
=
m.
m.
From t = 6 to t = 8 the distance traveled is
|f(8) - f(6)|
The total distance traveled in the first eight seconds is
(f) Because of what we learned in parts (d) and (e), we need to calculate the distances traveled during the time intervals [0, 2], [2, 6], [6, 8]
separately.
The distance traveled in the first 2 seconds is
(larger t-value). Thus, the particle is at rest after
m.
0, that is,
) or when both factors are negative (t <
and t >
m.
t₂
s (smaller t-value)
=
). Thus the particle moves in
It moves backward (in the negative direction) when](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd8f73ea3-8b9d-41a5-8140-775c62a3c253%2Fa3bfdec3-5066-4f33-998b-a0ded6231014%2F3p8i5z5_processed.png&w=3840&q=75)
Transcribed Image Text:(a) The velocity function is the derivative of the position function.
s = f(t) = t³ - 12t² + 36t
ds
dt
v(t) = - 3t² - 24t+ 36
=
=
(b) The velocity after 2 s means the instantaneous velocity when t = 2, that is
v(2):
= $12-2²
ds
dt
t=2
The velocity after 3 s is
v(3) = 3(3)² - 24(3) + 36 =
t=6
s=0
that is,
(c) The particle is at rest when v(t)
3t² - 24t+ 36 = 3(t² − 8t + 12) = 3(t − 2)(t - 6) =
(smaller t-value) or t =
and this is true when t =
and after
t=0
S=0
=
= 3(2)2 24(2) + 36 =
|f(2)=f(0)| =
|f(6) - f(2)=
=
(d) The particle moves in the positive direction when v(t) ?
3t² - 24t+ 36 = 3(t-2)(t-6) ? ⁰.
=
This inequality is true when both factors are positive (t >
the positive direction in the time intervals t < t₁
t₁ = [
<t<
|= t₂.
(e) Using the information from part (d) we make a schematic sketch in the figure of the motion of the particle back and forth along a line (the s-axis).
=
m/s.
s (larger t-value).
=
t=2
s=32
From t = 2 to t = 6 the distance traveled is
m/s.
=
m.
m.
From t = 6 to t = 8 the distance traveled is
|f(8) - f(6)|
The total distance traveled in the first eight seconds is
(f) Because of what we learned in parts (d) and (e), we need to calculate the distances traveled during the time intervals [0, 2], [2, 6], [6, 8]
separately.
The distance traveled in the first 2 seconds is
(larger t-value). Thus, the particle is at rest after
m.
0, that is,
) or when both factors are negative (t <
and t >
m.
t₂
s (smaller t-value)
=
). Thus the particle moves in
It moves backward (in the negative direction) when
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