v(t)=3t^2-2t-1 at any time t>=0 x(2)=5?

Question

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

Answer 1

integrate first...
x(t) = t³ - t² - t + C
x(2) = 5 ... then 8 - 4 - 2 + C = 5
C = 3
note also that v is zero when t = -1/3 & 1
... that means it changed direction at t = 1

to get the total distance... we need
|x(1) - x(0)| + |x(3) - x(1)| = |2 - 3| + |18 - 2| = 1 + 16 = 17


§

Answer 2

I'll suppose (since you don't say) that x is distance and v is speed, so v = dx/dt, where t is time.

Then to get an expression for x we integrate the expression for v with respect to time, so we get:

x = t^3 - t^2 - t + C

At t=2, x=5, so we can calculate C = 3

At t = 0, x = 3, and at t = 3, x = 18

So for the interval 0<=t<=3, total distance travelled is 18-3, or 15.

To get to this answer, you don't actually need to know C, as it cancels out, so the information that x(2)=5 is superfluous. Is there another part of this question that you are not telling us, which needs that information?

Addition:
D'Oh - Charlie above is right - I have calculated displacement, not total distance travelled! So it is a distance of 1 for 0<=t<=1 and 16 for 1<=t<=3, making a total of 17.

I'll read the question more carefully in future.

Answer 3

Is v(t)=3t^2-2t-1 velocity and x(t) distance
If so then x(t) = t^3 - t^2 -t +C now x(2) = 5 gives C = 3

A change in direction occurs when velocity changes sign ie at v=0 and this holds when t = 1 [and -1/3 which we discard since t>=0

the distance is then measured from t=0 to t=1 then from t=1 to t=3 and taking positive values added

Answer 4

So God has also opened Customer Support department. Great!!!!

Answer 5

I =s(t) = ∫3t² - 2t - 1 dt
I = s(t) = t³ - t² - t between t = 0 and t = 3
s(t) = 27 - 9 - 3
s(t) = 15 units is distance travelled.

Answer 6

dx/dt=v
dx=vdt

x=3 t^3/3 - 2 t^2/2-t+C

x= t^3 - t^2-t+C
5=8-4-2+C
C=3

Try to find the rest.
Study your math or physics book.
search google for examples.