Question

vectors help please.

Answer 1

An objects motion is describes by a position vector r=x(t)i + y(t)j, where x(t)=-0.31t^2+7.2t+28 and y(t)=0.22t^2-9.1t+30. find the velocity of the object after t=5s.

Answer 2

it's asking for the velocity so i don't have to take the derivative of anything right? just plug in the 5 into the equations? @surry99

Answer 3

if i do it that way i get r=56.5i-10j

Answer 4

please you have to calculate the subsequent derivatives first:
\[\frac{ dx }{ dt },\frac{ dy }{ dt }\]

Answer 5

to get V you need to take dr/dt

Answer 6

so it's going to end up as r=4.2i-6.9j ??

Answer 7

how did you arrive at that?

Answer 8

uhhh... i took the derivative of the x(t) and y(t) functions then plugged it into the r function

Answer 9

not quite....if you use Pythagoras:
V = (Vx^2 + Vy^2) ^1/2

Answer 10

@zephyr141 please calculate your derivatives, at time t=5

Answer 11

\[x(t)=-0.3t^2+7.2+28\]\[x'(t)=-0.6t+7.2\]\[x'(5)=-0.6(5)+7.2=4.2i\]for the x. then y i get\[y'(5)=0.44(5)-9.1=-6.9j\]then i plugged that into\[r=4.2i-6.9j\]

Answer 12

that's right!

Answer 13

so plugging into @surry99 i get 8.077 units/s

Answer 14

That is V = 4.2i - 6.9...it is not r since you took the derivatives

Answer 15

hmm.... so to find r do i just leave the equations alone and not take the derivatives?

Answer 16

oops typo...
V=4.2i−6.9j

Answer 17

@zephyr141 please note that @surry99 is right!
it is \[\frac{ dr }{ dt }\]

Answer 18

If you wanted r only then you just sub in the expressions and then r is a function of time.
Then you took the derivatives of both sides with respect to t to give your result:
V=4.2i−6.9j

Answer 19

@zephyr141 using the formula written by @surry99 you will get 8.08

Answer 20

so just\[r=(-0.3t^2+7.2t+28)i+(0.22t^2-9.1t+30)j\]

Answer 21

yes that is r then you take the derivative of both side wrt to time and that gives you the velocity expression at any time.

Answer 22

\[\frac{ dr }{ dt }=\left( \frac{ dx }{ dt },\frac{ dy }{ dt } \right)=\]
\[=\left( -0.6t +7.2,0.44t-9.1 \right)\]

Answer 23

So V =( -.6t +7.2)i + (.44t-9.1)j
Now you can plug in any time t and find the velocity in terms of the i and j unit vectors.
The final step is to get the magnitude of the velocity which is where Pythagoras comes in since the i and j component vectors are at right angles to each other.
V = (Vx^2 + Vy^2)^1/2 where Vx = dx/dt and Vy = dy/dt

Answer 24

ok so the answer is one of the four here:
A. r(5)=4.2i-6.9j
B. r(5)=-0.62i+0.44j
C. r(5)=56.3i-10j
D. r(5)=2.1i-3.5j

Answer 25

please, note that you have v(5), not r(5), so A.

Answer 26

These are just the position vectors they have provided not velocity.
So:
r(t) =(−0.3t2+7.2t+28)i+(0.22t2−9.1t+30)j

for t = 5
r(5) = (-.3(5)^2 + 7.2(5) + 28) i + (.22(5)^2 - 9.1(5) + 30) j
r(5) = 56.5 i + -10 j

Answer 27

r(t) =(−0.31t2+7.2t+28)i+(0.22t2−9.1t+30)j
(you had -.3t^2 it is -.31t^2 in the first term)

r(5) = 56.3 i + -10j

Answer 28

alrighty

Answer 29

Does it all make sense to you now?

Answer 30

yeah. i'm getting there. i'm going to reread that section of the book now.

Answer 31

ok, basically what you have in this problem is you start with:
r(t) - position as a function of time
if you know want the velocity as a function of time, you take the derivative of both sides wrt to time.
they could also ask for acceleration as a function of time so you would then take the derivative wrt to time of both sides of the velocity expression
So to review:
r(t) =(−0.31t2+7.2t+28)i+(0.22t2−9.1t+30)j
v(t) = dr/dt = (-.62t + 7.2) i + (.44t - 9.1) j
a(t) = dv/dt = -.62i + .44 j

Answer 32

note that acceleration components are actually independent of time

Answer 33

ok so the answer is one of the four here:
A. r(5)=4.2i-6.9j
B. r(5)=-0.62i+0.44j
C. r(5)=56.3i-10j
D. r(5)=2.1i-3.5j

One other very important point, the notation that is being used in these answers all represent position vectors ...r(t).
If they are asking for velocity, the notation should change such as v(t)
if they are asking for acceleration , the notation should change to a(t)

Answer 34

If a particular force has a potential energy U(x)=3x-6x^3 find an expression for the force.
so what i did here was to take the derivative again like the previous question i asked and got\[U'(x)=F(x)=3-18x^2\]

Answer 35

yes

Answer 36

Sorry, I think if U(x) is conservative, then, we can write:
\[F _{x}=-\frac{ \partial U }{ \partial x }=18x ^{2}-3\]

Answer 37

Note the negative sign @zephyr141

Answer 38

i see it. so does that mean 18x2-3 is the correct one

Answer 39

@zephyr141 if U(x) is conservative!

Answer 40

if U(x) is conservative then F= 18x^2-3