## anonymous one year ago The velocity function, in feet per second, is given for a particle moving along a straight line. v(t) = t^3 − 9t^2 + 20t − 12, 1 ≤ t ≤ 7 Find the total distance that the particle travels over the given interval.

1. ybarrap

Distance is integral of velocity because velocity is the derivative of distance over time: $$d=\int_1^7v(t)dt=\int_1^7\left (t^3 − 9t^2 + 20t − 12\right )~dt$$ Can you evaluate this integral?

2. anonymous

yesssss I did I don't know why my answer is wrong

3. IrishBoy123

that will give you displacement think about something like: $$\int_{1}^{7}|v(t)|dt$$

4. IrishBoy123

|dw:1444090723561:dw|

5. anonymous

yes then it changes at (t-1) (t-2) (t-6)

6. IrishBoy123

it's in the interval $$1 ≤ t ≤ 7$$ so the $$2 \le t \le 6$$ is the bit that needs to be thought about |dw:1444091029721:dw|

7. anonymous

okay.. what needs to be thought about it? I did all of the integrals

8. ybarrap

@IrishBoy123 is correct. Evaluate the "absolute" value of v(t) and split into three intervals rather than just a blanket 1-7. Otherwise, the way I did it, you will be subtracting distance rather than adding distance for that negative part -- not what you want.

9. anonymous

yeah i split it into my intervals

10. IrishBoy123

and you got ... ?

11. anonymous

-518 which doesn't make sense

12. IrishBoy123

$\color{red} -\int_2^6 (t^3 − 9t^2 + 20t − 12 ) dt = 32$

13. anonymous

I'm going to attach a picture of my work

14. anonymous

15. IrishBoy123

that's really hard to read $$\int \; t^3-9 t^2+20 t-12 \; dt = \dfrac{t^4}{4}-3 t^3+10 t^2-12 t \; [+C]$$ $$\int_{1}^{2} \; |v(t)| \; dt = \dfrac{3}{4}$$ $$\int_{2}^{6} \; |v(t)| \; dt = 32$$ $$\int_{6}^{7} \; |v(t)| \; dt = \dfrac{53}{4}$$