Question

The velocity of an object moving along a straight line is v(t) = t^2-10 t+16. Find the displacement over the time interval [1, 7]. Find the total distance traveled by the object.

Answer 1

her is a clue how do you get from velocity to position?

Answer 2

integrate it

Answer 3

1/3t^3 - 5t^2 + 16t

Answer 4

yes

Answer 5

so should i just put s(7) - s (1)

Answer 6

that will give me displacement or distance ?

Answer 7

That would give you displacement.

Answer 8

you can think of displacement as like net distance

Answer 9

ok how to find total distance ?

Answer 10

the same way you would find area instead of net area

Answer 11

\[v(t)=t^2-10t+16\]
When is v positive and when is v negative on the interval 1 to 7?

Answer 12

2 to 5 ?

Answer 13

if you factor v we get (x-2)(x-8)

------|------|-----
2 8
positive negative positive


So you will have to split your integral into two integrals
one going from 1 to 2 and the other going from 2 to 7

Answer 14

\[|\int\limits_{1}^{2}v dt | +| \int\limits_{2}^{7} v dt |\]

Answer 15

ok so i integrate vt and get st then do s(2) -s(1) + s(7) - s(2) ?

Answer 16

|s(2)-s(1)|+|s(7)-s(2)| <--correction

Answer 17

oh ok thanks

Answer 18

You might think this is useful:
http://saylor.org/site/wp-content/uploads/2011/04/DISPLACEMENT-vs.pdf

Answer 19

is displacement -ve ?

Answer 20

i got -30 for displacement

Answer 21

that's good

Answer 22

but how can displacement be -ve ?

Answer 23

I'm pretty that signifies the direction

Answer 24

the negative part

Answer 25

you are moving backwards

Answer 26

and the total distance i got is -7.27

Answer 27

am i doing something wrong

Answer 28

yeah distance shouldn't be negative

Answer 29

did you do
|s(2)-s(1)|+|s(7)-s(2)|
?

Answer 30

yes

Answer 31

you do no it is impossible for absolute value of something to be negative right?

Answer 32

know*

Answer 33

yes so its 7.273

Answer 34

so let me see if i can help you with the arithmetic

Answer 35

that website say tha we should subtract the - part from +.....and we are adding both integrals

Answer 36

\[s(t)=\frac{t^3}{3}-10 \frac{t^2}{2} +16t+C\]
\[s(t)=\frac{t^3}{3}-5t^2+16t+C\]

\[|s(2)-s(1)|=|(\frac{2^3}{3}-5(2)^2+16(2))-(\frac{1^3}{3}-5(1)^2+16(1)|\]
\[|s(7)-s(2)|=|(\frac{7^3}{3}-5(7)^2+16(7))-(\frac{2^3}{3}-5(2)^2+16(2))|\]
After you computer these
Add them

Answer 37

compute

Answer 38

the absolute value around those differences mean you need to take the absolute value

Answer 39

you are suppose to be adding a positive to a positive

Answer 40

we have + and - so should it be s(+) - s(-)

Answer 41

\[|\frac{8}{3}-20+32-\frac{1}{3}+5-16|+|\frac{343}{3}-245+112-\frac{8}{3}+20-32|\]

Answer 42

oh ok so answer is 40.65 or 41 ?

Answer 43

\[|\frac{7}{3}+12-11|+|\frac{335}{3}-133-12|\]
\[|\frac{7}{3}+1|+|\frac{335}{3}-145|\]
\[|\frac{10}{3}|+|\frac{335-435}{3}|\]
\[|\frac{10}{3}|+|\frac{-100}{3}|=\frac{10}{3}+\frac{100}{3}=\frac{110}{3}\]

Answer 44

you do know that if x<0 then |x|=-x
and when x>=0 then |x|=x

that is why they put negatives in front of the negative parts
and positives and front of the positive
it is the definition of absolute value

Answer 45

oh ok so displacement is -30 and distance is 36.6

Answer 46

i would leave it as 110/3
instead of 36.7

Answer 47

unless you are asked to round to the neartest 10th

Answer 48

then it would be 36.7

Answer 49

also do you understand the arithmetic now?

Answer 50

yea thanks and displacement part is -30 right ?

Answer 51

that is what we got

Answer 52

cool thanks for help

Answer 53

np