Finding the distance from the velocity equation.

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MooreLikeMike

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Hi everyone,

I have the question: "Suppose the velocity of a car is given by v(t) = 1/(2t+1)+(4t+3)^2 feet per second. How far does the car travel in the first two seconds?"

I know that v(t) is the derivative of the equation for the distance traveled, I'll just call it D(t). So to find D(t), I need to find the indefinite integral of v(t). Then after finding D(t) = 1/2ln(abs(2t+1))+1/12(4t+3)^3+C, I should set "t" equal to zero and solve for "C" to find D(t).

But where I'm confused is because I know if the question were to ask to find "how far does the car travel in 3 to 6 seconds", I would find the definite integral from 3 to 6 and that would be the answer. But since the question that I'm given asks to find the distance in the first 2 seconds, I not sure if I'm supposed to find the definite integral from 0 to 2, or if I should just plug 2 in for "t" and solve it that way.

Because if I find the definite integral from 0 to 2, "C" will be eliminated, but if I don't, "C" will still be apart of the answer.
 
First it is customary to call the distance formula s(t) (I have no idea why). However if you want to call it D(t) that is perfectly fine. You have D(t) so you want to compute D(2) - D(0) (or D(6) - D(3)) or integrate v(t)dt from 0 to 2.

Remember, D(2) tells you where you are after 2 seconds, but that does not mean you travelled D(2) feet! You need to subtract D(0) from D(2) to find the total distance travelled.

Think about how you do this with you car. You arrive somewhere and your odometer reads 20,318 miles. Does this mean during this trip you travelled 20,318 miles? Probably not. You need to subtract the starting reading on the odometer to know how far you travelled for this trip!
 
Jomo said:
First it is customary to call the distance formula s(t) (I have no idea why). However if you want to call it D(t) that is perfectly fine. You have D(t) so you want to compute D(2) - D(0) (or D(6) - D(3)) or integrate v(t)dt from 0 to 2.

Remember, D(2) tells you where you are after 2 seconds, but that does not mean you travelled D(2) feet! You need to subtract D(0) from D(2) to find the total distance travelled.

Think about how you do this with you car. You arrive somewhere and your odometer reads 20,318 miles. Does this mean during this trip you travelled 20,318 miles? Probably not. You need to subtract the starting reading on the odometer to know how far you travelled for this trip!
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Thank you, Jomo! That makes a lot of sense!
 
Jomo said:
First it is customary to call the distance formula s(t) (I have no idea why)...
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The vector [math]\vec{s}[/math] is defined to be the displacement. I presume it isn't usually called [math]\vec{d}[/math] in order to avoid problems with the Leibniz notation for the derivative, [math]\dfrac{d}{dt}[/math]. The choice as to the actual letter "s" is a mystery to me as well.

-Dan
 
Someone here must know why the letter s is used for displacement. Please enlighten us!
 
displacement ...

[math]\int_{t_0}^{t_f} v(t) \, dt[/MATH]
distance traveled ...

[MATH]\int_{t_0}^{t_f} |v(t)| \, dt[/math]
note distance traveled [MATH]\ge[/MATH] displacement
 
From a Google search I found that some say that "S" stands for the Latin word "spatium", which means distance or space.
 
topsquark said:
The vector [math]\vec{s}[/math] is defined to be the displacement. I presume it isn't usually called [math]\vec{d}[/math] in order to avoid problems with the Lorentz notation for the derivative, [math]\dfrac{d}{dt}[/math]. The choice as to the actual letter "s" is a mystery to me as well.

-Dan
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Leibniz?
 
Jomo said:
Ooo, I missed that one. Good catch! Do you think that deserves some corner time?
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I believe that under the Lorenz transformations it is impossible to spend any time in the corner
 
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