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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.
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.

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ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1Distance 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yesssss I did I don't know why my answer is wrong

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2that will give you displacement think about something like: \(\int_{1}^{7}v(t)dt\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444090723561:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes then it changes at (t1) (t2) (t6)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2it'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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay.. what needs to be thought about it? I did all of the integrals

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1@IrishBoy123 is correct. Evaluate the "absolute" value of v(t) and split into three intervals rather than just a blanket 17. Otherwise, the way I did it, you will be subtracting distance rather than adding distance for that negative part  not what you want.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i split it into my intervals

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0518 which doesn't make sense

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\[\color{red} \int_2^6 (t^3 − 9t^2 + 20t − 12 ) dt = 32\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm going to attach a picture of my work

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2that's really hard to read \(\int \; t^39 t^2+20 t12 \; dt = \dfrac{t^4}{4}3 t^3+10 t^212 t \; [+C]\) \(\int_{1}^{2} \; v(t) \; dt = \dfrac{3}{4}\) \(\int_{2}^{6} \; v(t) \; dt = 32\) \(\int_{6}^{7} \; v(t) \; dt = \dfrac{53}{4}\)
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