Differential equations

[hamza4best]

The thing is that I'm not good at making differential equations out of word problems.And I've some wod problems which I tried solving but failed.I'm posting the problems here not just because I need a quick solution but I'm also hoping that some one here can help me out in uderstanding and making differential equations out of any word problem.

1. A particle starting with an initial velocity of 25m/s moves in a straight line through a resisting medium which decreases the velocity of the particle at a constant rate of 10m/s each second.How far will the particle travel before coming to rest?

2. The brakes on a certain car can stop the car in 200 feet from a speed of 55 miles/hour.Assume that when the brakes are applied the car has a constant negative acceleration. (a) How much time in seconds is required to bring the car to a stop from 55 miles/hour? (b) If the car is brought to a stop from 55 miles/hour how far will it have moved by the time its speed is reduced to 25 miles/hour?

3. From the top edge of a building 20 meters high a stone is throne vertically upward with an initial velocity of 30 meters/second (a) In how many seconds will the stone strike the ground? (b) How high will the stone rise? (c) How fast will the stone be falling when it hits the ground?

4. A balloon is rising at the constant rate of 10 feet/second and is 100 feet from the ground at the instant when the astronaut drops his binoculars. (a) How long will it take the binoculars to strike the ground? (b) With what speed will the binoculars strike the ground?

5. A projectile is fired vertically upward by a cannon with an initial velocity of vo meters per second.At what speed will the projectile be moving when it returns and strikes the hapless cannoneer(Neglect air resistance)

6. A balloon is rising vertically at a constant rate of 1 meters per second and has reached an altitude of 8 meters at the instant when an assistant on the ground directly under the balloon attempts to toss the astronauts binoculars up to her.What is the minimum velocity with which the binoculars should be thrown straight upward if they are released from the assistant's hand at an altitude of 2 meters?

7. Suppose that a particle P moves along the s axis with a constant acceleration a.Let vo be the velocity of the particle when t=0 and let so be its coordinate when t=0.Show that (a) v=at+vo (b) s=1/2at^2 +vot + so

8. A stone is dropped from a height of h meters with zero initial velocity and it hits the ground T seconds later.Show that h=4.9T^2

9. Consider the differential equation dy/dt =k(A-y)(B+y) for the formation of trypsin in the small intestine.Assuming that A>B determing the time t at which trypsin is being formed most rapidly.

 

[stapel]

...I tried solving but failed. I'm...hoping that some one here can help me out in uderstanding....

Please reply showing the results of your efforts, so the helpers can see where you're getting lost. In this way, they'll be much better placed to help you understand what's going on. Thank you!

 

[HallsofIvy]

All of those problems involve constant acceleration. Do you know how to write acceleration as a derivative?

 

[hamza4best]

All of those problems involve constant acceleration. Do you know how to write acceleration as a derivative?

Yes acceleration is the time rate of change of velocity i.e dv/dt
I'm having problem making equations through the problems

 

[DrPhil]

Yes acceleration is the time rate of change of velocity i.e dv/dt
I'm having problem making equations through the problems

I'll show you (1) as an example. Do whatever you can, starting from what is given, one step at a time - and ALWAYS keep track of the UNITS!

a = dv/dt = -10 m/s^2

Initial v_0 = 25 m/s

Time to stop = t = -v_0/a = (25 m/s)/(10 m/s^2) = 2.5 s
[note that the units tell you what arithmetic to do!]

average velocity = (v_0 + v_final)/2 = (25 m/s + 0)/2 = 12.5 m/s

distance = v_avg*t = (12.5 m/s)(2.5 s) = 31.25 m

If you show us your work on some of the others, we can guide you where you get stuck.

 

[Deleted member 4993]

The thing is that I'm not good at making differential equations out of word problems.And I've some wod problems which I tried solving but failed.I'm posting the problems here not just because I need a quick solution but I'm also hoping that some one here can help me out in uderstanding and making differential equations out of any word problem.

1. A particle starting with an initial velocity of 25m/s moves in a straight line through a resisting medium which decreases the velocity of the particle at a constant rate of 10m/s each second.How far will the particle travel before coming to rest?

Another way:

You want to find distance traveled given the initial and final velocities and acceleration.

a = dv/dt = dv/ds * ds/dt = v * dv/ds ← differential equation

a * ds = v * dv

$\displaystyle a * \displaystyle \int_0^s ds \ = \ \int_{v_i}^{v_f} v * dv$

Now finish it.... Incidentally, this is Galileo's third equation of motion.

2. The brakes on a certain car can stop the car in 200 feet from a speed of 55 miles/hour.Assume that when the brakes are applied the car has a constant negative acceleration. (a) How much time in seconds is required to bring the car to a stop from 55 miles/hour? (b) If the car is brought to a stop from 55 miles/hour how far will it have moved by the time its speed is reduced to 25 miles/hour?

3. From the top edge of a building 20 meters high a stone is throne vertically upward with an initial velocity of 30 meters/second (a) In how many seconds will the stone strike the ground? (b) How high will the stone rise? (c) How fast will the stone be falling when it hits the ground?

4. A balloon is rising at the constant rate of 10 feet/second and is 100 feet from the ground at the instant when the astronaut drops his binoculars. (a) How long will it take the binoculars to strike the ground? (b) With what speed will the binoculars strike the ground?

5. A projectile is fired vertically upward by a cannon with an initial velocity of vo meters per second.At what speed will the projectile be moving when it returns and strikes the hapless cannoneer(Neglect air resistance)

6. A balloon is rising vertically at a constant rate of 1 meters per second and has reached an altitude of 8 meters at the instant when an assistant on the ground directly under the balloon attempts to toss the astronauts binoculars up to her.What is the minimum velocity with which the binoculars should be thrown straight upward if they are released from the assistant's hand at an altitude of 2 meters?

7. Suppose that a particle P moves along the s axis with a constant acceleration a.Let vo be the velocity of the particle when t=0 and let so be its coordinate when t=0.Show that (a) v=at+vo (b) s=1/2at^2 +vot + so

8. A stone is dropped from a height of h meters with zero initial velocity and it hits the ground T seconds later.Show that h=4.9T^2

9. Consider the differential equation dy/dt =k(A-y)(B+y) for the formation of trypsin in the small intestine.Assuming that A>B determing the time t at which trypsin is being formed most rapidly.

.

 

[hamza4best]

I'll show you (1) as an example. Do whatever you can, starting from what is given, one step at a time - and ALWAYS keep track of the UNITS!

a = dv/dt = -10 m/s^2

Initial v_0 = 25 m/s

Time to stop = t = -v_0/a = (25 m/s)/(10 m/s^2) = 2.5 s
[note that the units tell you what arithmetic to do!]

average velocity = (v_0 + v_final)/2 = (25 m/s + 0)/2 = 12.5 m/s

distance = v_avg*t = (12.5 m/s)(2.5 s) = 31.25 m

If you show us your work on some of the others, we can guide you where you get stuck.

I tried solving (7)

heres what i've got

dv/dt=a

dv=a dt
v=at + c1
vo when t=0

vo=0=c1
c1=vo

similarly
ds=(at+c1)dt
S=at^2/2 +c1t +c2

So when t=0

so=0 + 0 + c2

so=c2

s=1/2at^2 + vot + so

is this correct?

In (9)

we have dy/dt =k(A-y)(B+y)
is it a good idea to multily the terms?
dy/dt=k(AB+Ay-By-y^2)
dy/dt=ABk+Aky-Bky-ky^2

(Bky-Aky+ky^2)dy=(ABk)dt

1/2(Bky^2) - 1/2(Aky^2)+(1/3ky^3)=ABkt

then what? how do I find t?

Now (3)

acc=-g
dv/dt=-g
dv=-g dt
v=-gt + c1
30=-g(0) + c1
30=c1
v=-gt +30
ds/dt=(-gt+30)
s=-gt^2/2 + 30t + c2
s=-4.9t^2 + 30t +c2
20=0 + 0 + c2
c2=20

when it will strike the ground s will become zero so

4.9t^2-30t-20=0

applying the quadratic formula i get t=6.73s

now for finding velocity

v=+gt + 30 (I took +g as now we have to find v when it will move downward)

v=95.95m/s

for S again I took +g

S=+4.9(6.73)^2 + 30(6.73) +20
S=443.835 feet

These are the one's I had a slight idea how to solve in the rest of these I need your help

 

[hamza4best]

I tried solving (7)

heres what i've got

dv/dt=a

dv=a dt
v=at + c1
vo when t=0

vo=0=c1
c1=vo

similarly
ds=(at+c1)dt
S=at^2/2 +c1t +c2

So when t=0

so=0 + 0 + c2

so=c2

s=1/2at^2 + vot + so

is this correct?

In (9)

we have dy/dt =k(A-y)(B+y)
is it a good idea to multily the terms?
dy/dt=k(AB+Ay-By-y^2)
dy/dt=ABk+Aky-Bky-ky^2

(Bky-Aky+ky^2)dy=(ABk)dt

1/2(Bky^2) - 1/2(Aky^2)+(1/3ky^3)=ABkt

then what? how do I find t?

Now (3)

acc=-g
dv/dt=-g
dv=-g dt
v=-gt + c1
30=-g(0) + c1
30=c1
v=-gt +30
ds/dt=(-gt+30)
s=-gt^2/2 + 30t + c2
s=-4.9t^2 + 30t +c2
20=0 + 0 + c2
c2=20

when it will strike the ground s will become zero so

4.9t^2-30t-20=0

applying the quadratic formula i get t=6.73s

now for finding velocity

v=+gt + 30 (I took +g as now we have to find v when it will move downward)

v=95.95m/s

for S again I took +g

S=+4.9(6.73)^2 + 30(6.73) +20
S=443.835 feet

These are the one's I had a slight idea how to solve in the rest of these I need your help

In (2) I'm not too sure about this but I gave it a go

s=0 when t=0
vi=55miles/hour when t=0
s=200 feet when t=?

so firstly converted the 55 miles/hour into feets/sec and I got 80.67feet/sec

dv/dt=a

v=at +c1

ds/dt=at+c1 dt

s=at^2/2 + c1t +c2

0=0+0+c2
so c2=0

vi=80.67feet/sec when t=0

80.67=c1

s=1/2(at^2) + 80.67t
200=0+80.67t

so t=2.47sec

Now here I get stuck..I don't know how to solve (b)

 

[DrPhil]

#7

7. Suppose that a particle P moves along the s axis with a constant acceleration a.Let vo be the velocity of the particle when t=0 and let so be its coordinate when t=0.Show that (a) v=at+vo (b) s=1/2at^2 +vot + so

I tried solving (7)

heres what i've got

dv/dt=a

dv=a dt
v=at + c1
vo when t=0

vo=0=c1
c1=vo

similarly
ds=(at+c1)dt
S=at^2/2 +c1t +c2

So when t=0

so=0 + 0 + c2

so=c2

s=1/2at^2 + vot + so

is this correct?

YES!! You have proved the Equation of Motion with constant acceleration, Very good.

 

[DrPhil]

#9

9. Consider the differential equation dy/dt =k(A-y)(B+y) for the formation of trypsin in the small intestine.Assuming that A>B determing the time t at which trypsin is being formed most rapidly.

In (9)

we have dy/dt =k(A-y)(B+y)
is it a good idea to multily the terms?
dy/dt=k(AB+Ay-By-y^2)
dy/dt=ABk+Aky-Bky-ky^2 = ABk + k(A - B)y - ky^2

(Bky-Aky+ky^2)dy=(ABk)dtl ..... NO, dy/(k(A - B)y - ky^2) = (ABk)dt
................................................or, dy/[(A-y)(B+y)] = k dt

then what? how do I find t?

This might be the hardest one of the bunch, because it is asking a totally different question. You are given a formula for the rate of change, and the question is when that rate has its maximum. The problem may be wrong - you should be setting d/dt[dy/dt] = 0, but what you have for dy/dt is NOT a function of t, so the 2nd derivative would be identically zero. But, maybe . . .

OPTIONS: Differentiate dy/dt with tespect to y, and find the value of y when the rate is maximum.

Integrate the differential equation using partial fractions. That will give a (logarithmic function of y) = kt.
If you know y, you can then find t.

 

[DrPhil]

2. The brakes on a certain car can stop the car in 200 feet from a speed of 55 miles/hour.Assume that when the brakes are applied the car has a constant negative acceleration. (a) How much time in seconds is required to bring the car to a stop from 55 miles/hour? (b) If the car is brought to a stop from 55 miles/hour how far will it have moved by the time its speed is reduced to 25 miles/hour?

In (2) I'm not too sure about this but I gave it a go

s=0 when t=0
vi=55miles/hour when t=0
s=200 feet when t=?

so firstly converted the 55 miles/hour into feets/sec and I got 80.67feet/sec OK

dv/dt=a

v=at +c1

ds/dt=at+c1

s=at^2/2 + c1t +c2

0=0+0+c2
so c2=0

vi=80.67feet/sec when t=0

80.67=c1

s=1/2(at^2) + 80.67t Two unknowns, a and t
200=0+80.67t............Can't set a=0 and still stop the car!
What is the average speed, slowing linearly from 80.67 ft/s down to 0?

so t=2.47sec.....Off by a factor of 2

Now here I get stuck..I don't know how to solve (b)

Once you know the total time to stop, you can solve for a. Use the equation
v=at +c1
to find out how long it takes to decrease speed by 30 m/h.

 

[DrPhil]

#3

3. From the top edge of a building 20 meters high a stone is throne vertically upward with an initial velocity of 30 meters/second (a) In how many seconds will the stone strike the ground? (b) How high will the stone rise? (c) How fast will the stone be falling when it hits the ground?

Now (3)

acc=-g
dv/dt=-g
dv=-g dt
v=-gt + c1
30=-g(0) + c1
30=c1
v = -gt + 30 --> v(t) = -gt + (30 m/s)
ds/dt=(-gt+30)
s=-gt^2/2 + 30t + c2
s=-4.9t^2 + 30t +c2 -->
s(t) = - (4.9 m/s^2)t^2 + (30 m/s)t + (20 m)
this is the equation of motion you need - ALWAYS include UNITS!
20=0 + 0 + c2
c2=20

when it will strike the ground s will become zero so

4.9t^2 - 30t - 20=0

applying the quadratic formula i get t=6.73s yes

now for finding velocity

v=+gt + 30 (I took +g as now we have to find v when it will move downward)
.................BUT if you switch signs, then v0 would be negative!!
.................Probably better to stay with + being upward and - being downward

v = 95.95 m/s NO, v_final = -35.94 m/s or v_final = 35.94 m/s downward

for S again I took +g Bad idea

S=+4.9(6.73)^2 + 30(6.73) +20 ... S(6.73) has to be 0 .. that is how you found 6.73 s
S=443.835 feet

To get the maximum height, you have to know where the stone is when the velocity has dropped to zero - that is at the peak of its trajectory.
v(t) = - (9.8 m/s^2)t + (30 m/s)
v(t) = 0 --> t = ??
Then find s(t) for that value of t.

 

[hamza4best]

To get the maximum height, you have to know where the stone is when the velocity has dropped to zero - that is at the peak of its trajectory.
v(t) = - (9.8 m/s^2)t + (30 m/s)
v(t) = 0 --> t = ??
Then find s(t) for that value of t.

what about (4),(5),(6) and (8)
can you give some starting steps. I'm unable to do these.

 

[hamza4best]

Two unknowns, a and t
...........Can't set a=0 and still stop the car!
What is the average speed, slowing linearly from 80.67 ft/s down to 0?

I didn't get what you meant
avg speed= 80.67ft/s -vf/2=40.35ft/sec..is that what you meant?
and then?

 

[hamza4best]

what about (4),(5),(6) and (8)
can you give some starting steps. I'm unable to do these.

for (8)

I know that
vi=0 then t=0

and then t becomes T

but how to use vi?

h(t)=g

so d^2h/dt^2=g

h=gt^2/2 + c

h=4.9t^2 + c

 

[DrPhil]

4. A balloon is rising at the constant rate of 10 feet/second and is 100 feet from the ground at the instant when the astronaut drops his binoculars. (a) How long will it take the binoculars to strike the ground? (b) With what speed will the binoculars strike the ground?

5. A projectile is fired vertically upward by a cannon with an initial velocity of vo meters per second.At what speed will the projectile be moving when it returns and strikes the hapless cannoneer(Neglect air resistance)

6. A balloon is rising vertically at a constant rate of 1 meters per second and has reached an altitude of 8 meters at the instant when an assistant on the ground directly under the balloon attempts to toss the astronauts binoculars up to her.What is the minimum velocity with which the binoculars should be thrown straight upward if they are released from the assistant's hand at an altitude of 2 meters?

8. A stone is dropped from a height of h meters with zero initial velocity and it hits the ground T seconds later.Show that h=4.9T^2

In question (3) you found the equation of motion for a specific s0 and v0. By this time you should have it memorized. If not, go through the process again of integrating acceleration to get velocity and integrating velocity to get position. Let the initial conditions be that at t=0, v = v0 and s = s0. Then

s(t) = s0 + v0 t + (1/2)a t^2

Memorize it. Use it. Make it a part of your everyday conversation. Read about it. See the movie. Remember also that v = ds/dt, and a = dv/dt. Wear it on your forehead, write it on your cuff, and etch it on an urn. Truth is beauty, beauty truth. That's all ye know and all ye need to know.