This is from my engineering mechanics bookI'm not that familiar with limits yet in integral calculus, and I'm not that good with advanced calculus yetand in the image is a solved problembut I don't understand the steps on how it got to the answercan you explain the steps to me and I think there were shortcuts made in the solution can you give me the longer version or tell me any fundamental formulas I should already know when solving this kind of problem, so I can understand how the book derived to the answer. A small projectile is fired vertically downward into a fluid medium withan initial velocity of 60 m/s. Due to the drag resistance of the fluid theprojectile experiences a deceleration of a = (-0.4v³) m/s², where v is inm/s. Determine the projectile's velocity and position 4 s after it is fired.SOLUTIONCoordinate System. Since the motion is downward, the positioncoordinate is positive downward, with origin located at O, Fig. 12-3.Velocity. Here a = f(v) and so we must determine the velocity as afunction of time using a = dv/dt, since this equation relates v, a, and t.(Why not use v = vo + act?) Separating the variables and integrating,with = 60 m/s when t = 0, yieldsdv(+↓)dta =VScom/s-0.42²601-0.4(2)110.8²=x = {[(60)²V-0.4v³==V12² 601(60)²dt+ 0.8t=t-0= 11]¯'^²}xm/sHere the positive root is taken, since the projectile will continue tomove downward. When t = 4 s,v = 0.559 m/sAns.Fig. 12-3 Here the positive root is taken, since the projectile will continue tomove downward. When t = 4 s,v = 0.559 m/sAns.Position. Knowing v = f(t), we can obtain the projectile's positionfrom v = ds/dt, since this equation relates s, v, and t. Using the initialconditions = 0, when t = 0, we have(+↓)When t = 4 s,v=S =1[² ds = [(60)²SoS =ds1dt (60)²0.4=210.8 (60)²+ 0.8t1-1/2+ 0.8t dt+ 0.8t81] ¹2/001{[(60)² + 0.81]1/²s = 4.43 m-160mAns.

Given data: vi=60 m/sa=-0.4v3 m/s2t=4 sec Need to determine the position and velocity at t=4 sec.

Answered: A small projectile is fired vertically…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

This is from my engineering mechanics book
I'm not that familiar with limits yet in integral calculus, and I'm not that good with advanced calculus yet

and in the image is a solved problem

but I don't understand the steps on how it got to the answer

can you explain the steps to me and I think there were shortcuts made in the solution can you give me the longer version or tell me any fundamental formulas I should already know when solving this kind of problem, so I can understand how the book derived to the answer.

A small projectile is fired vertically downward into a fluid medium with
an initial velocity of 60 m/s. Due to the drag resistance of the fluid the
projectile experiences a deceleration of a = (-0.4v³) m/s², where v is in
m/s. Determine the projectile's velocity and position 4 s after it is fired.
SOLUTION
Coordinate System. Since the motion is downward, the position
coordinate is positive downward, with origin located at O, Fig. 12-3.
Velocity. Here a = f(v) and so we must determine the velocity as a
function of time using a = dv/dt, since this equation relates v, a, and t.
(Why not use v = vo + act?) Separating the variables and integrating,
with = 60 m/s when t = 0, yields
dv
(+↓)
dt
a =
V
Scom/s-0.42²
60
1
-0.4
(2)
1
1
0.8²
=
x = {[(60)²
V
-0.4v³
=
=
V
1
2² 60
1
(60)²
dt
+ 0.8t
=t-0
= 1
1]¯'^²}x
m/s
Here the positive root is taken, since the projectile will continue to
move downward. When t = 4 s,
v = 0.559 m/s
Ans.
Fig. 12-3

Here the positive root is taken, since the projectile will continue to
move downward. When t = 4 s,
v = 0.559 m/s
Ans.
Position. Knowing v = f(t), we can obtain the projectile's position
from v = ds/dt, since this equation relates s, v, and t. Using the initial
conditions = 0, when t = 0, we have
(+↓)
When t = 4 s,
v=
S =
1
[² ds = [(60)²
So
S =
ds
1
dt (60)²
0.4
=
2
1
0.8 (60)²
+ 0.8t
1-1/2
+ 0.8t dt
+ 0.8t
81] ¹2/0
0
1
{[(60)² + 0.81]1/²
s = 4.43 m
-
1
60
m
Ans.

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