Question

a particle moving on the x axis has position x(t)= 2t^3+3t^2-36t+40

find the total distance traveled by the particle during the first 3 seconds

The answer is 61

keep getting |x(1)-x(0)|+|x(3)-x(1)|=35

I got the zeros by factoring the derivative of x(t) (v(t)).

Answer 1

\[x(t)= 2t^3+3t^2-36t+40\]
\[=\int\limits_0^3 2t^3+3t^2-36t+40 \text d t\]\[=\left.\frac{t^4}{2}+\frac 3 2t^3-18t^2+40t\right|_0^3\]\[=\frac{(3)^4}{2}+\frac 3 2(3)^3-18(3)^2+40(3)\]
\[=\frac {81}{2}+\frac 3 2 \times 27 -18\times 9 +120\]

Answer 2

except that doesn't give 61, there is probably a few error in my working

Answer 3

This is in the related rate section I can't use that lol

Answer 4

I'll give you a medal anyways :)

Answer 5

the second them the 3/2 bit should have been 3/3=1

Answer 6

I have to find the zero's of the velocity because that is when the "particle" changes direction then calculate the distance

Answer 7

zeros*

Answer 8

oh, so to find the zeroes take the first derivative of x(t)

Answer 9

yeah but I did that I just keep getting 35? idk what I did wrong

Answer 10

\[x(t)= 2t^3+3t^2-36t+40 \]
\[\dot x=\frac{\text d x }{\text d t}=6t^2+6-36=6(t^2-5)\]

Answer 11

looks like it is turning at t= √5

Answer 12

6t*

Answer 13

not 6 lol

Answer 14

?

Answer 15

x'(t)= 6t^2+6t-36

Answer 16

oh yeh

Answer 17

hehe

Answer 18

\[x(t)=2t^3+3t^2−36t+40\]
\[x'=6t^2+6t−36=6(t^2+t-6)=6(t-2)(t+3)\]

Answer 19

so turning at t= 2 and t= 3

Answer 20

thats what I did wrong! thanks!

Answer 21

t=-3 i mean

Answer 22

so you should break the integral into
\[\int_0^2x(t)\text d t+\int_2^3x(t)\text d t\]