Integrals and motion: a(t) = 10 sin(t) + 3 cos(t), and....

[scrum]

Hello, I'm just starting math 2, which means we're doing integration. I'm having some trouble with my homework, and I think it's because they're conceptual questions.

I'm quite frustrated as I thought I had a good grasp of the subject and I'm getting answers to questions that I think are right but that are wrong, so it's almost worse because i have a wrong understanding of the material rather than no understanding.

I'm going to post some of the problems. Some of them I have not attempted yet. After posting I have another class and after that I will read my book, have another crack at them and then come back here.

First One

Let s(t) denote the position of a particle at time t, and let v and a be the velocity and acceleration respectively. The particle is moving according to the data
a(t) = 10sin(t) + 3cos(t) , s(0)=8, s(pi/2)= -10

For this one I integrated it twice.
v(t) = -10cos(t) + 3sin(t) + C

s(t) = -10sin(t) - 3cos(t)+Ct+D

and then plugged in pi/2 and zero.
Here is my problem..

s(pi/2) = -10 + C(pi/2)+D
s(0) = -3 +D

Now S(0) is meant to be 8 so that makes D = 11
But Then I need to make
s(pi/2) = -10
s(pi/2) = -10 + C(pi/2) + 11

so then s(pi/2) = 1 + c(pi/2)

C = - 22/pi

I thought I had it just then but I dont and it's wrong and i've got class so i will work on it when i get back and then check again.

 

[stapel]

Let s(t) denote the position of a particle at time t, and let v and a be the velocity and acceleration respectively. The particle is moving according to the data
a(t) = 10sin(t) + 3cos(t) , s(0)=8, s(pi/2)= -10

Okay; we have the acceleration equation and two data points for the position. But we don't have any actual question here. What are you supposed to be doing with this information?

I will guess from what you posted that you are supposed to try to find the equation for the position. If so, then use the fact that a(t) = v'(t) and v(t) = s'(t), where v(t) is the velocity function.

. . . . .a(t) = 10 sin(t) + 3 cos(t)

. . . . .v(t) = int [ 10 sin(t) + 3 cos(t) ] dt

. . . . . . . .= -10 cos(t) + 3 sin(t) + C

...where "C" is the constant of integration. Continuing, we have:

. . . . .s(t) = int [ -10 cos(t) + 3 sin(t) + C ] dt

. . . . . . . .= -10 sin(t) - 3 cos(t) + Ct + D

...where "D" is the constant of integration. Plugging in the two data points, we get:

. . . . .s(0) = 0 - 3 + D = 8

. . . . .D = 11

. . . . .s(pi/2) = -10 + (pi/2)C + 11 = -10

. . . . .(pi/2)C + 11 = 0

. . . . .(pi/2)C = -11

. . . . .C = (-11)(2/pi) = -22/(pi)

Then:

. . . . .s(t) = -10 sin(t) - 3 cos(t) - (22/pi)t + 11

However, you say that this is not the correct answer, which implies that finding the position function was not what you were supposed to do...?

Please reply with clarification. Thank you!

Eliz.

 

[scrum]

Yea I realized what I was doing wrong.

The pi/2 was a 2pi.

Which makes me kind of happy because it meansmy concept understanding is good but i just misread the question.

However I have some more that are confusing me. Often I get them as I type my work, so here goes...

An object is thrown upwards with initial velocity w meters per second from a point d meters above the ground. Assume that the acceleration due to gravity is -9.8 meters per second squared, and let v(t) and s(t) denote the velocity and position of the object above the ground, respectively, after t seconds.

(a) Determine a function that describes the velocity v of the object after time t.

v(t) = [quote:14f88xfc]w-9.8t I got this answer, it correct

(b) Determine a function that describes the position s of the object after time t.

s(t) = [/quote:14f88xfc]wt-(4.9t^2)+d I got this answer, it also correct[/quote]

(c) Use (a) and (b) to obtain a formula that relates the velocity v of the object with w, s, and d. Your formula should have the form V^2 = [/quote]

For this part I think i'm misunderstanding what they want, I'm sure i could do it if i understood what they were asking me to express in terms of what. How is V^2 different from V?

 

[Deleted member 4993]

Yea I realized what I was doing wrong.

The pi/2 was a 2pi.

Which makes me kind of happy because it meansmy concept understanding is good but i just misread the question.

However I have some more that are confusing me. Often I get them as I type my work, so here goes...

An object is thrown upwards with initial velocity w meters per second from a point d meters above the ground. Assume that the acceleration due to gravity is -9.8 meters per second squared, and let v(t) and s(t) denote the velocity and position of the object above the ground, respectively, after t seconds.

(a) Determine a function that describes the velocity v of the object after time t.

v(t) = [quote:15i3k1hy]w-9.8t ..............................................(1)

solve for 't'..............................................(1)

I got this answer, it correct

(b) Determine a function that describes the position s of the object after time t.

s(t) =

wt-(4.9t^2)+d

replace 't' from (1)

I got this answer, it also correct[/quote:15i3k1hy]

(c) Use (a) and (b) to obtain a formula that relates the velocity v of the object with w, s, and d. Your formula should have the form V^2 = [/quote]

For this part I think i'm misunderstanding what they want, I'm sure i could do it if i understood what they were asking me to express in terms of what. How is V^2 different from V?[/quote]