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 Jom/s v= 60 m/s-0.4v-³ 1.4 (12) 20 -0.4 60 1 0.8 -0.4v-³ {[(60)² + = dt =t-0 1 (60)² + 0.8t 81]¹/²} m 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

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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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
Transcribed Image Text: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.
Transcribed Image Text: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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