Calculus help plz?

Question

Can some one plz help me figure out this question, I really dont understand how to do it.

A particle moves along the x-axis so that its velocity at any time 't' greater than or equal to zero is given by..... v(t)=3t^2-2t-1.
The position x(t) is 5 for t=2

Find the total distance traveled by the particle from time t=0 until time t=3.

Thank you very much for your help, I really appreciate it!

Answer 1

v(t) = 3t² - 2t - 1

Integrate to get x(t).

x(t) = t³ - t² - t + c
x(2) = 8 - 4 - 2 + c = 5
2 + c = 5
c = 3

x(t) = t³ - t² - t + 3

Net distance traveled between t = 0 and t = 3 is:

x(3) - x(0) = (27 - 9 - 3 + 3) - (0 - 0 - 0 + 3) = 18 - 3 = 15

But the partical didn't travel in the same direction the whole time.

v(t) = 3t² - 2t - 1 = 0
(3t + 1)(t - 1) = 0
t = -1/3, 1
Ignore -1/3 because it is negative.

Direction was negative for t = 0 to t = 0. Then positive for the remainder. Total distance traveled then is:

|x(1) - x(0)| + |x(3) - x(1)| = |2 - 3| + |18 - 2| = 1 + 16 = 17

The total distance traveled is 17.

Answer 2

The total traveled distance is 17.
First off, find the intergration of the v(t) which would give the equation to the position. That comes out to be t^3 - t^2 - t (+-) c
To find c you plug in 2 for t in this equation and have it equal to 5. In the end you get 5=2. So c must equal 3. So the final equation is t^3 - t^2 -t + 3. Then you just plug in 0,1,2,3 for t and you get x(0)=3, x(1)=2, x(2)=5, x(3)=18. 3-2=1, 5-2=3, 18-5=13. (13+3+1=17)

The answer can't be 15 because it ask for total distance. You have to account for the total movement the particle did. It started at 3 and went back 1 step then went forward 3 steps then went forward again 13 steps. so 13+3+1=17. Therefore the answer is 17 not 15, nor 20, or any other number.

Answer 3

You have the velocity equation, v(t). To get the distance travelled during a time span, you integrate the velocity equation to get the distance equation, x(t).

Integration of v(t) is t^3 - t^2 - t + c. This is x(t).

Use the fact that x(2) = 5 to solve x(t) for c:
2^3 - 2^2 - 2 + c = 5
8 - 4 - 2 + c = 5
2 + c = 5
c = 5 - 2
c = 3

So x(t) = t^3 - t^2 - t + 3.

You then calculate x(3) - x(0) to get the answer.

x(3) = 27 - 9 - 3 + 3 = 18
x(0) = 0 - 0 - 0 + 3 = 3.

So the answer is 18-3 = 15

Answer 4

Take the integration of v(t) to get x(t) and it is x(t)=t^3-t^2-t+C
Then substitute and you get 5=2^3-2^2-2+C
Which means C equals 3.
The position equation is x(t)=t^3-t^2-t+3
So then plug in t=0 and t=3 into the postion equation.
You get: x(0)=3, which means the initial position of the particle
was 3.
x(3)=27-9-3+3 which simplifies to x(3) equals 18.
Find the difference between 18 and 3 to get the answer.
The answer is 15.
Hope this helps!

Answer 5

first, take the integral to find the equation for the position of the particle...

x(t) = int (t = 0, t) v(t) = t^3-t^2-t+C

To find C, plug in the given position at time t=2:

5 = 2^3-2^2-2+C = 2+C ---------> C = 3

now plug in time t=3 to find the final position....

x(t=3) = 3^3-3^3-3+3 = 18

and plug in t=0 to find the initial position...

x(t=0) = 0-0-0+3 = 3

So the total distance travelled is 18-3 = 15.




edit: mich 1 is right, but my solution is more complete. it's good practice if you're learning calc. ;-)

Answer 6

First u should find when the particle is at rest.So v(t)=0.
3t^2-2t-1=0
(3t+1)(t-1)=0
t=-1/3 or t=1
rejected
To get distance, u should integrate v(t), it gives, t^3-t^2-t+c.
x=5,t=2,
5=(2^3)-(2^2)-2+c
c=3
x(t)=t^3-t^2-t+3
t=1, x(t)=2
t=3,x(t)=18
Thus, total distance = 18+2 = 20.
Hope this helps.

Answer 7

integrate V=3t^2-2t-1
to get t^3-t^2-t and then you plug in 3 into it to get 15, which is your answer

Reasoning: you don't need to do all those complicated steps that the other guys are talking about. What you are trying to find is really the area under the curve of the velocity function because velocity*time=distance: therefore you integrate to get the area under the curve which is the distance :D

Answer 8

15

Answer 9

they are soliciting for the spinoff of y with respect to x. do you should use limits or are you able to employ the regulations for derivation? For limits it really is going to seem as if: dy/dx = decrease (h -> 0) [ f(x+h) - f(x) ] / [(x+h)-x] Then basically plug on your numbers.

Answer 10

Runs screaming cowering in the corner sucking thumb! Kudos to you for doing calculus Is scares the bejesus out of me!