Question

A point moves in a straight line so that its distance at time t from a fixed point of the line is 8t - 3t^2. What is the total distance covered by the point between t=1 and t=2?

Answer 1

\[\int\limits_{1}^{2}8t-3t ^{2} dx\]

Answer 2

5

Answer 3

A particle moves in a straight line with velocity v(t) = t^2. How far does the particle move between the times t=1 and t=2? same type of problem, but for some reason I'm getting a strange answer. I integrated to get s(t) and then tried to follow the process above, but...

Answer 4

\[\int_1^2t^2dt\] yes?

Answer 5

yes

Answer 6

why is the answer strange?

Answer 7

oh... I integrated again because it was asking for position? So I integrated to get s(t) and then again to use the interval?

Answer 8

i get
\[2\tfrac{1}{3}\]

Answer 9

you want "distance traveled" which is the integral provided the integrand is positive over than interval

Answer 10

ok. So I don't need to find s(t)?

Answer 11

you found s(t) when you took the antiderivative of
\[t^2\]
\[s(t)=\frac{t^3}{3}-1\]

Answer 12

no that is wrong.
\[s(t)=\frac{t^3}{3}-\frac{1}{3}\]

Answer 13

maybe this is not clear. the derivative of the position function is the velocity and you know the velocity is
\[v(t)=t^2\] so the position must be
\[s(t)=\frac{t^3}{3}+C\] right?

Answer 14

yes

Answer 15

you also know that you are starting at
\[t=1\] so the position there is 0

Answer 16

(you haven't gone anywhere yet, that is your initial spot)

Answer 17

Doesn't the first problem give you the distance function

Answer 18

got it

Answer 19

Zarkon: yes

Answer 20

so put
\[s(1)=0\] and solve for C you get
\[s(1)=\frac{1^3}{3}+C=0\] and so
\[C=-\frac{1}{3}\]

Answer 21

I'm following

Answer 22

that is one explanation of why when you integrate is is
\[F(b)-F(a)\]

Answer 23

your position function is now known to be
\[s(t)=\frac{t^3}{3}-\frac{1}{3}\] so if you want to know where you are at t = 2 just plug it in and get
\[s(2)=\frac{2^3}{3}-\frac{1}{3}\]

Answer 24

which is another way of saying if you want to know how far you have gone compute
\[\int_1^2t^2dt\]

Answer 25

but the problem doesnt actually state that it starts at t=1... that is just the beginning of the interval. Couldn't it already be in motion?

Answer 26

you are starting at t = 1. think of it this way: if you are traveling at 60 miles per hour it and you want to know how far you travel from 1 pm to 2 pm it doesn't matter when you started your trip, or what mile marker you are at at 1 pm

Answer 27

oh ok. I understand all of it now. Thank you for your patience

Answer 28

yw. but maybe we better look at question 1 again because i don't think it is right

Answer 29

it looks like the first one you are told your position. it says distance is given by
\[8t - 3t^2\], not velocity

Answer 30

ok

Answer 31

so at 1 pm you are at mile marker 5, and at 2 you are at mile marker

Answer 32

dang thing keep jumping around

Answer 33

ok this is not trivial. at 1 you are at 5, and at
t=4/3
you are at
\[5\tfrac{1}{3}\]
and then you start heading back and at 2 you are at 4

Answer 34

so you went 1/3 forward and 2 back for a total distance traveled of
\[2\tfrac{1}{3}\]

Answer 35

that is not one of the answer choices though. If we want to know the total distance why wouldn't we use s(t)?

Answer 36

what are the choices for #1?

Answer 37

5/3?

Answer 38

maybe i messed up. zarkon?

Answer 39

5/3 is an option. Can you explain your process Zarkon?

Answer 40

5->51/3=1/3
51/3->4=4/3
1/3+4/3=5/3

Answer 41

i hope i made a calculation error, not a conceptual one

Answer 42

i just used your numbers

Answer 43

oh right, damng

Answer 44

right from 5 and a third to 4 is 1 and a third. doh

Answer 45

one can also do this

\[f(t)=8t-3t^2\]

distance is

\[|f(1)-f(4/3)|+|f(4/3)-f(2)|=5/3\]

Answer 46

not one third forward, 2 back. one third forward and one and one third back

Answer 47

hope it is clear, even though i screwed up, that we just found the vertex of the parabola, said it was increasing until 4/3 and then decreasing until 2

Answer 48

I'm really confused where all the thirds are coming from, sorry... would one of you mind starting from the beginning with a concise explanation?

Answer 49

i messed up so i will let zarkon explain. probably be better

Answer 50

lol...you were doing fine

Answer 51

as satellite noticed the position function goes away and then back between the values of t=1 and t=2

Answer 52

so we need to find when it switches direction

Answer 53

that is at it's maximum (located at t=4/3) use the vertex formula

Answer 54

the position at time 1 was f(1)=8(1)-3(1^2)=5
and the position at 4/3 was f(4/3)=5+1/3

Answer 55

what's the vertex formula? I'm blanking

Answer 56

for \[ax^2+bx+c\]

its
\[x=\frac{-b}{2a}\]

Answer 57

we had \[-3t^2+8t\]

\[t=\frac{-8}{2(-3)}=\frac{4}{3}\]

Answer 58

continuing .... so the distance moving away was (5+1/3)-5=1/3

Answer 59

but t is going to 2, not 4/3

Answer 60

4/3 is on the way from 1 to 2

Answer 61

so now we are ate f(4/3)=5+1/3
and we are moving to f(2)=4
that distance is (5+1/3)-4=1+1/3=4/3

Answer 62

so the total distance is the 1/3 out plus the 4/3 back

1/3+4/3=5/3

Answer 63

btw although i generally eschew this explanation, the derivative is the velocity. your derivative is
\[-6t+8\] so you can see it is positive (you are going forward) when

\[t<\frac{4}{3}\] and negative (you are going back) when
\[t>\frac{4}{3}\]

Answer 64

so you can alway find the vertex by taking the derivative, setting it to zero and solving. of course you will alway get
\[-\frac{b}{2a}\] in any case

Answer 65

turned out to be a bit of a problem didn't it?

Answer 66

good stuff :)

Answer 67

\[\checkmark\]

Answer 68

wow. ok, thanks to both of you. The only thing is I'm not allowed to use a calc for this problem, so how would I necessarily know the graph changes? Or would I go with the velocity method?

Answer 69

it is a parabola .. it will always change direction...use the vertex formula to find where it changes

Or use calculus

Answer 70

oh ok! So if the vertex was not in the interval, , would I just solve it the way the first guy did it?

Answer 71

no....

you would evaluate the function at the two values and take the difference

Answer 72

right ugh sorry, summer makes me forget all my math skills