Calculating Distance Traveled Using Velocity Function and Integration

In summary, when given the equation v = 8t - 3t^2 and the starting point of the body as P, the distance of the body from P at t=3 can be found by integrating the equation and taking the difference between s(t) and s(0). The arbitrary constant C can be ignored, resulting in the final answer of 9 meters.
  • #1
FaraDazed
347
2

Homework Statement



if [itex]v = 8t - 3t^2[/itex] and the body is at P

find where the body is from P when t=3

The Attempt at a Solution



I am very new to calculus and have just been taught basics of differentiation and integration.

I know that when you integrate there is an arbitrary constant so I have go this far (below).

[itex]∫ ( v = 8t - 3t^2 ) dt = 4t^2 - t^3 + C[/itex]

I am not sure what to do with the constant, is it ignored and thus the answer is as below

[itex]s = 4t^2 - t^3[/itex]
[itex]s = (4 \times 9) - 27[/itex]
[itex]s = 36 - 27[/itex]
[itex]s = 9[/itex]metres ?

Or do I have to do something with the constant?
 
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  • #2
Normally at t=0, s=0 and v=0. So your constant C will work out to be 0. So what you did is correct.
 
  • #3
FaraDazed said:

Homework Statement



if [itex]v = 8t - 3t^2[/itex] and the body is at P

find where the body is from P when t=3

The Attempt at a Solution



I am very new to calculus and have just been taught basics of differentiation and integration.

I know that when you integrate there is an arbitrary constant so I have go this far (below).

[itex]∫ ( v = 8t - 3t^2 ) dt = 4t^2 - t^3 + C[/itex]

I am not sure what to do with the constant, is it ignored and thus the answer is as below

[itex]s = 4t^2 - t^3[/itex]
[itex]s = (4 \times 9) - 27[/itex]
[itex]s = 36 - 27[/itex]
[itex]s = 9[/itex]metres ?

Or do I have to do something with the constant?

The distance traveled is s(t) - s(0), from which C drops out.

RGV
 
  • #4
rock.freak667 said:
Normally at t=0, s=0 and v=0. So your constant C will work out to be 0. So what you did is correct.

Ray Vickson said:
The distance traveled is s(t) - s(0), from which C drops out.

RGV

OK thanks for the help :)
 

1. What is the purpose of integrating a velocity function?

The purpose of integrating a velocity function is to determine the displacement of an object over a certain time period. This is done by finding the area under the velocity curve, which represents the distance traveled by the object.

2. How is the integral of a velocity function related to the derivative of a position function?

The integral of a velocity function is the inverse operation of the derivative of a position function. This means that the derivative of a position function gives the velocity at a certain point, while the integral of a velocity function gives the total displacement over a certain time period.

3. Can the integral of a velocity function be negative?

Yes, the integral of a velocity function can be negative. This occurs when the velocity function is negative for a certain time period, indicating that the object is moving in the opposite direction. The negative integral value represents the displacement in the opposite direction.

4. How is the average velocity of an object calculated using the integral of a velocity function?

The average velocity of an object can be calculated by dividing the total displacement, obtained from the integral of the velocity function, by the total time period. This gives the average rate of change of position over the given time period.

5. Can the integral of a velocity function be used to determine the acceleration of an object?

Yes, the integral of a velocity function can be used to determine the acceleration of an object. The derivative of the velocity function gives the acceleration, and the integral of the acceleration function gives the change in velocity over a certain time period.

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