Question

A particle moving in a straight linehas a velocity...

Answer 1

@timo86m

Answer 2

it will be

(v((1/4)*pi)-v((1/6)*pi))/((1/4)*pi-(1/6)*pi)

Answer 3

is that the answer or is that the first step?

Answer 4

that is the answer plug it in :)

Answer 5

basically the formula is this

(f(x2)-f(x1))/(x2-x1)

Answer 6

what do i plug in? 2sint-1? and if so, do i plug it in for both of the v's?

Answer 7

well yes and one for each

like this hold on

Answer 8

(2*sin((1/4)*pi)-1-2*sin((1/6)*pi)+1)/((1/4)*pi-(1/6)*pi)

Answer 9

ok on sec let me solve it-my answer will be in decimal form right?

Answer 10

http://www.wolframalpha.com/input/?i=%282*sin%28%281%2F4%29*pi%29-1-2*sin%28%281%2F6%29*pi%29%2B1%29%2F%28%281%2F4%29*pi-%281%2F6%29*pi%29

Answer 11

this is how it should look like :)

Answer 12

wow. ok ill work on that haha. can we move on to the next problem?

Answer 13

ok

Answer 14

To find acceleration or rate of change of velocity, you need to do \[( v(\Pi/4)-v(\Pi/6) )/ \Pi/4-\Pi/6\]

Answer 15

Basically change in y/change in x

Answer 16

for 50 set it to 0=v(t)

Answer 17

2sint+1=0?

Answer 18

btw are you in degrees or radians? yes that is exactly it :)

Answer 19

im not exactly sure since it doesnt specify... and what do i do after i set it = to 0?

Answer 20

solve for t

Answer 21

but doesnt it say what are the first 2 values?

Answer 22

t=arcsin(-1/2) i think

Answer 23

0s will accur at regular intervals :)

Answer 24

every 2pi i think :)

Answer 25

im sorry but you lost me :P i got up to sint=-1/2

Answer 26

that could work too
graph it online and see where it is zero :)

Answer 27

im not supposed to use a calc but ok i will

Answer 28

you need a calc to find arcsin i think.

Answer 29

what is arcsin? is that the same as sin of theta?

Answer 30

it is the inverse of sin.

Answer 31

like if you had 1+x=y

you would

subtract to solve

it is to sin what subtraction is to addition.

Answer 32

so i basically do sin^-1(-1/2)
and when i do i get -.523... as my t. am i doing it right so far?

Answer 33

i get -(1/6)*Pi for first 0

Answer 34

well ok wolframalpha says t is -pi/6. so once i have this what do i do?

Answer 35

usually it repeats every 2 pi

Answer 36

oh do i just do what you said earlier which is to add 2pi?

Answer 37

lol youre answering my questions before i can type them :) so i guess the second value would be 11pi/6

Answer 38

i think so let me double check by graphing it

Answer 39

it is wrong that every 2 pi thing

https://www.google.com/search?rlz=1C1CHFX_enUS470US470&aq=f&sugexp=chrome,mod=0&sourceid=chrome&ie=UTF-8&q=2*sin(x)%2B1

here is the real answer :)

Answer 40

but the first one is -pi/6?

Answer 41

3.6 and 5.7 well it is wrong cuzz i think the q says t>0

Answer 42

ughh

Answer 43

https://www.google.com/search?rlz=1C1CHFX_enUS470US470&aq=f&sugexp=chrome,mod=0&sourceid=chrome&ie=UTF-8&q=2*sin(x)%2B1

just look at the graph

Answer 44

ya i see i just dont know how to show my work

Answer 45

it is a tought one to do BUt remember I told you it was every 2 pi well 3.6 accurs 2pi after -pi/6 :) as for 5.7 i am working on that :)

Answer 46

you can try function manipulation and shift it to the left or right like this

v(t+c)

Answer 47

-pi/6 + 2pi is 5.7

Answer 48

i dont get how to get the 3.6

Answer 49

3.6 was gotter graphically and also you can do it by -pi/6+2 pi

Answer 50

if you allow to use a ti 83 you can use that to graph and find the zeros aka roots.

Answer 51

ya ill just use my calc cause i cant think of any other way.

Answer 52

lol i know i can but i am not feeling so well lol.

Answer 53

haha its fine. are u up to doing 51 and 52 or no?

Answer 54

for 52 use trig id thatsin (x+.5*pi)=cos(x)

so it is .5pi=a

Answer 55

51 just plug in

Answer 56

plug in what?

Answer 57

51


5.38791281 to 2.41831093

plug into your 1+5*cos(t/2) function

Answer 58

where did those 2 #s come from?

Answer 59

i pluged in 1 and 10 for 1+5*cos(t/2)

Answer 60

ok and on 52 where did the .5 come from?

Answer 61

oh that is just a trig id sin(x+.5pi)=cos (x)

Answer 62

oh ive nvr heard of that b4

Answer 63

so what would the value of x be or atleast how do u fin it?

Answer 64

.5pi :)

Answer 65

i thought you said that was the value of a??

Answer 66

it is

Answer 67

it is communative property either x or a can be .5pi

Answer 68

and it is also the value of x?

Answer 69

yup
it is like saying

x+a=1

and lets say either x or a can be 1 and the other 0 :)

Answer 70

oh ok

Answer 71

if x is .5pi then a is a

if a is .5pi then x is x

Answer 72

got it. well i got everything but 50 lol but thanks so much!

Answer 73

well just draw sin x

stretch it by a factor of 2 and shift it up by 1.

Answer 74

ok! thanks :)

Answer 75

what is the period on number 51?

Answer 76

i mean on 50

Answer 77

2 pi