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Wnh5001

Senior member
Dec 1, 2005
408
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Hello ATOT'ers i am really in need of help, hw is due soon, and ive been working on this problem and other complex( for me at least) problems. ok here is the question.

The acceleration of a particle along an x axis is a = 5.0t, with t in seconds and a in meters per second squared. At t = 2.0 s, its velocity is +16 m/s. What is its velocity at t = 6.0 s?

Equations u might need.
V=Vo + at
X-Xo = Vot + 1/2at^2
v^2=Vo^2 + 2a(X-Xo)
X-Xo=1/2(Vo+v)t
X-Xo=vt-1/2at^2
any help is appreciated..
 
C

Compnewbie01

Senior member
Aug 8, 2005
603
4
81
Okay I oversimplified it way too much. Since A(t) = 5t, I think you would use the antiderivative(sp?) for V(t).
 
W

Wnh5001

Senior member
Dec 1, 2005
408
0
76
yea i dont know calc yet, its a calc based physics, my advisors and the course description says u can take calc concurrently with this course... and im have an extremely difficult time, while everyone else is using integrals and deriatives to rape all their problems.
 
H

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: Compnewbie01
Would that mean that the initial velocity (Vo) is 6 m/s and that it would be going 36 m/s at t = 6 seconds? Since it was 16 m/s at t = 2 seconds, it accelerated 10 m/s during that time.
Click to expand...

That's exactly what I got.

V = Vo + at
Sub in the point you know:
16 = Vo + 5 (2)

Vo = 6.

Sub in second point:

V = 6 + 5 (6) = 36m/s.

Alternate solution:
a(t) = 5
Integrate to find v(t)

v(t) = 5t + c
v(2) = 16 --> c = 6

v(6) = 5(6) + 6 = 36m/s.
 
Soccerman06

Soccerman06

Diamond Member
Jul 29, 2004
5,830
5
81
Originally posted by: hypn0tik
Originally posted by: Compnewbie01
Would that mean that the initial velocity (Vo) is 6 m/s and that it would be going 36 m/s at t = 6 seconds? Since it was 16 m/s at t = 2 seconds, it accelerated 10 m/s during that time.
Click to expand...

That's exactly what I got.

V = Vo + at
Sub in the point you know:
16 = Vo + 5 (2)

Vo = 6.

Sub in second point:

V = 6 + 5 (6) = 36m/s.

Alternate solution:
a(t) = 5
Integrate to find v(t)

v(t) = 5t + c
v(2) = 16 --> c = 6

v(6) = 5(6) + 6 = 36m/s.
Click to expand...

That sounds more right, but Im too busy studying to really do any work (copying someones hw ;))
 
W

Wnh5001

Senior member
Dec 1, 2005
408
0
76
36 m/s is a no go...one more try, this online hw gives us five tries, there is no mistake in wording of problem as i copy and pasted..
 
C

Compnewbie01

Senior member
Aug 8, 2005
603
4
81
I feel embarrassed. This question seems extremely easy and I am having second thoughts about 36 m/s. The antiderivative of A(t) = 5t is V(t) = 2.5x^2.
On the other hand, I don't see why 36 m/s doesn't work.
 
T

talyn00

Golden Member
Oct 18, 2003
1,666
0
0
Originally posted by: hypn0tik
Originally posted by: Compnewbie01
Would that mean that the initial velocity (Vo) is 6 m/s and that it would be going 36 m/s at t = 6 seconds? Since it was 16 m/s at t = 2 seconds, it accelerated 10 m/s during that time.
Click to expand...

That's exactly what I got.

V = Vo + at
Sub in the point you know:
16 = Vo + 5 (2)

Vo = 6.

Sub in second point:

V = 6 + 5 (6) = 36m/s.

Alternate solution:
a(t) = 5
Integrate to find v(t)

v(t) = 5t + c
v(2) = 16 --> c = 6

v(6) = 5(6) + 6 = 36m/s.
Click to expand...

looks correct to me
 
W

Wnh5001

Senior member
Dec 1, 2005
408
0
76
here is another problem not worth anything though, do it if u are bored and need to refresh your physics lol...>_>

"A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of 0.45 s, and the top-to-bottom height of the window is 2.50 m. How high above the window top did the flowerpot go?"

i already tried to answer it and used up my five total tries. my setup looked like 2.5=.225v-1/2(9.8m/s^2)(.225)^2 and solved for v, inputted back into some equation to find that displacment above the window = 4.09 m but it seems it was wrong, meh, maybe wrong sig figs.. >_<
 
W

Wnh5001

Senior member
Dec 1, 2005
408
0
76
exactly 36 u say? o_o.. maybe something wrong with this evil site that grades everything.
 
J

jchu14

Senior member
Jul 5, 2001
613
0
0
okay here's how you would approach it using simpel calc. I knwo you said you haven't learned it, but this makes it really simple, just get someone to teach you very simple differntiation and integration rules (product rule mainly) soon and you'll be all set for mechanics.

a=5t

integrate

v=(5/2)t^2+ C C is integration constant

you know
v(2)=16 ---> 16=(5/2)(2^2)+C ---> C=16-(5/2)*4 ----> C=6

solve for v(6)
v(t)=(5/2)t^2+6
v(6)=(5/2)*36+6
v(6)=96 m/s

i'll post a more through explanation later, i know you're in a rush
 
W

Wnh5001

Senior member
Dec 1, 2005
408
0
76
42. Is that your final answer? one try left, lol this feels like who wants to be a millionaire with no prize..
 
The J

The J

Senior member
Aug 30, 2004
755
0
76
Originally posted by: Wnh5001
Hello ATOT'ers i am really in need of help, hw is due soon, and ive been working on this problem and other complex( for me at least) problems. ok here is the question.

The acceleration of a particle along an x axis is a = 5.0t, with t in seconds and a in meters per second squared. At t = 2.0 s, its velocity is +16 m/s. What is its velocity at t = 6.0 s?

Equations u might need.
V=Vo + at
X-Xo = Vot + 1/2at^2
v^2=Vo^2 + 2a(X-Xo)
X-Xo=1/2(Vo+v)t
X-Xo=vt-1/2at^2
any help is appreciated..
Click to expand...

Let's see: The integral of a(t) is v(t), which would be v(t)=2.5t^2+C (there's always a constant since d/dx[C] = 0).

At t=2, you have v(2)= 2.5(2)^2 + C = 10+c = 16, I think. So C = 6 and v(t)=2.5t^2+6.

Now, just plug in t=6. v(6)= 2.5(36)+6= 96m/s.

That's what I got. Someone should check me, though.


EDIT: looks like jchu14 got it while I was typing.
 
J

jchu14

Senior member
Jul 5, 2001
613
0
0
Originally posted by: PottedMeat
Heh is this the UT Physics Homework Service?
Click to expand...

haha i was thinking that the whole time. I know UTHW all too well.

LONGHORNS REPRESENT! Aerospace Engineering c/o '08 (hopefully!) =)
 
silverpig

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Originally posted by: hypn0tik
Originally posted by: Compnewbie01
Would that mean that the initial velocity (Vo) is 6 m/s and that it would be going 36 m/s at t = 6 seconds? Since it was 16 m/s at t = 2 seconds, it accelerated 10 m/s during that time.
Click to expand...

That's exactly what I got.

V = Vo + at
Sub in the point you know:
16 = Vo + 5 (2)

Vo = 6.

Sub in second point:

V = 6 + 5 (6) = 36m/s.

Alternate solution:
a(t) = 5
Integrate to find v(t)

v(t) = 5t + c
v(2) = 16 --> c = 6

v(6) = 5(6) + 6 = 36m/s.
Click to expand...

A(t) is not 5. A(t) is 5t.
 
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