Need advice on structuring integration ("Given: an acceleration rate of sqrt(4t+1)....")

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SpockNard

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Given: an acceleration rate of sqrt(4t+1). The problem is to find the distance travelled in the interval from t=0>2. The thing that has me is that an initial velocity [v sub 0] is given as -4 1/3.
I can integrate and get the correct integral on a itself, and on that result [presumably v], and integrate v. But, these steps do not give me the book answer, which is 4 29/30. I have tried adding vsub0 to v, etc. How do I incorporate vsub0? Thanks!
 
SpockNard said:
Given: an acceleration rate of sqrt(4t+1). The problem is to find the distance travelled in the interval from t=0>2. The thing that has me is that an initial velocity [v sub 0] is given as -4 1/3.
I can integrate and get the correct integral on a itself, and on that result [presumably v], and integrate v. But, these steps do not give me the book answer, which is 4 29/30. I have tried adding vsub0 to v, etc. How do I incorporate vsub0? Thanks!
Click to expand...

Please reply showing what you tried, so we can see what's going on. Thank you!
 
find the indefinite integral of [imath]a(t)[/imath], using [imath]v_0[/imath] to determine the constant of integration of the velocity function.

then, recall distance traveled is [imath]\displaystyle \int_{t_0}^{t_f} |v(t)| \, dt[/imath]
 
SpockNard said:
Given: an acceleration rate of sqrt(4t+1). The problem is to find the distance travelled in the interval from t=0>2. The thing that has me is that an initial velocity [v sub 0] is given as -4 1/3.
I can integrate and get the correct integral on a itself, and on that result [presumably v], and integrate v. But, these steps do not give me the book answer, which is 4 29/30. I have tried adding vsub0 to v, etc. How do I incorporate vsub0? Thanks!
Click to expand...
I do get the book's answer, by doing what @skeeter said; and don't let the absolute value scare you too much -- just observe the sign of v over the indicated interval.

When we see your work, we'll be able to give more detailed advice, if it still needs correction.
 
SpockNard said:
Given: an acceleration rate of sqrt(4t+1). The problem is to find the distance travelled in the interval from t=0>2. The thing that has me is that an initial velocity [v sub 0] is given as -4 1/3.
I can integrate and get the correct integral on a itself, and on that result [presumably v], and integrate v. But, these steps do not give me the book answer, which is 4 29/30. I have tried adding vsub0 to v, etc. How do I incorporate vsub0? Thanks!
Click to expand...
In case you are still having difficulty understanding why your method didn't work, if your [imath]v_0[/imath] is negative and v is positive in the end, then the direction of motion changed. The total distance traveled will be first in the negative direction, then back to the origin, and then to the stopping point.

-Dan
 
just wondering about the mixed number representations for initial velocity and the solution … is that the form presented in the original text of the problem, or did you convert them from improper fractions?
 
I am older guy who likes to challenge brain by reviewing my own college calculus every several years. Crikey, I once knew this stuff cold. My Thomas Calculus and Analytic Geometry 4th Ed. has answers, not full solutions. Can’t find the latter anywhere so far. So, thanks for your patience. Don't have calculus notation keys.
Thomas 4th ed. 6.3 #12

Given a=sqrt(4t+1). vsub0= -(4 1/3). Find distance travelled t=0>2
Given answer 4 29/30

Integrate a for v
u=(4t+1) du/dt=4. dt=du/4
1/4*Integral sqrt u du = 1/4*(2/3)*u(3/2 power) =u(3/2 power)/6.
So, v=(4t+!)(3/2 power)/6

Using same strategy with U substitution,

Second integral of a= s = (4t+!)(5/2 power)/60

For t=2-0, = 243/60-1/60=242/60=121/30 = 4 1/30
Tried integrating v-vsub0
= integral of v (above) minus integral 4 1/3 = (4t+!)(5/2 power)/60 - 13t/3= [for t=0>2)26/3-0 =26/3
So, 121/30 - 260/30 = -139/30 = -4 19/30

As often, I am not grasping the format to structure the solution. Topsquark (Dan) may have the key, but I can't grasp how to combine the velocities correctly in the t=0>2 time frame.Thanks!
 
SpockNard said:
Don't have calculus notation keys.
Click to expand...
You can insert images of your work; or (eventually) learn a little LaTeX.

SpockNard said:
Integrate a for v
u=(4t+1) du/dt=4. dt=du/4
1/4*Integral sqrt u du = 1/4*(2/3)*u(3/2 power) =u(3/2 power)/6.
So, v=(4t+1)(3/2 power)/6
Click to expand...
All you're missing here is "+C". Now you need to use the initial value of v to find the value of C. (It is not [imath]v_0[/imath] itself; you'll plug [imath]t=0[/imath] into the equation and solve for C.)

The second integration has to wait until you've done that. In particular, you'll have to determine the sign of the resulting v over the given interval; it will turn out to be the same throughout, which is why I said not to worry too much about the absolute value.
 
Last edited:
SpockNard said:
Given a=sqrt(4t+1). vsub0= -(4 1/3). Find distance travelled t=0>2
Given answer 4 29/30

Integrate a for v
u=(4t+1) du/dt=4. dt=du/4
1/4*Integral sqrt u du = 1/4*(2/3)*u(3/2 power) =u(3/2 power)/6.
So, v=(4t+!)(3/2 power)/6
Click to expand...

Your formatting is a bit rough, but I'm pretty sure you have arrived at:

[imath]\qquad v(t) = \frac{1}{6}\, (4t + 1)^{3/2}[/imath]

But this does not take [imath]v_0[/imath] into account. Shouldn't it be more along the lines of the following?

[imath]\qquad v(t) = \frac{1}{6}\, (4t + 1)^{3/2} + C[/imath]

...where [imath]C = v_0[/imath]
 
stapel said:
Please reply showing what you tried, so we can see what's going on. Thank you!
Click to expand...

Dr.Peterson said:
You can insert images of your work; or (eventually) learn a little LaTeX.


All you're missing here is "+C". Now you need to use the initial value of v to find the value of C. (It is not [imath]v_0[/imath] itself; you'll plug [imath]t=0[/imath] into the equation and solve for C.)

The second integration has to wait until you've done that. In particular, you'll have to determine the sign of the resulting v over the given interval; it will turn out to be the same throughout, which is why I said not to worry too much about the absolute value.
Click to expand...
This is what had me befuddled: "Now you need to use the initial value of v to find the value of C." I've finally got it! Thanks for the LaTeX reference. I think I need to drop back a few chapters in Thomas. All the help is very much appreciated, but I'm afraid my Calculus lobe has too many lacunae to just forge ahead from where I left off. I do find Calculus fun, and I now am reminded that "C" is not just there for looks!
 
stapel said:
Your formatting is a bit rough, but I'm pretty sure you have arrived at:

[imath]\qquad v(t) = \frac{1}{6}\, (4t + 1)^{3/2}[/imath]

But this does not take [imath]v_0[/imath] into account. Shouldn't it be more along the lines of the following?

[imath]\qquad v(t) = \frac{1}{6}\, (4t + 1)^{3/2} + C[/imath]

...where [imath]C = v_0[/imath]
Click to expand...

C is not v(0) …

[imath]v(t) = \dfrac{1}{6} (4t+1)^{3/2} + C[/imath]

[imath]v(0) = -\dfrac{13}{3} = \dfrac{1}{6} [4(0)+1]^{3/2} + C \implies C = -\dfrac{9}{2}[/imath]

[imath] v(t) = \dfrac{1}{6}(4t+1)^{3/2} - \dfrac{9}{2}[/imath]

note [imath]v(t) = 0[/imath] at [imath]t=2[/imath] and is negative for [imath]0 \le t < 2[/imath]

therefore, distance traveled is [imath]\displaystyle \int_0^2 \dfrac{9}{2} - \dfrac{1}{6}(4t+1)^{3/2} \, dt[/imath]
 
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