Question

i always wanted to know how to solve these:

a particle moves according to the equation \(x = 10t^2\), where x is in meters and t is in seconds (a) Find the average velocity for the time interval from 2.00 s to 3.00 s

Answer 1

tale integral of x from 2 to 3

Answer 2

what about with no calclulus involved?

Answer 3

find x for t=2 and 3
calculate the difference
divide by time interval (here, 1 sec)
ans is diff/time

Answer 4

uhh would you do it step by step not flood the steps into me coz i cant visualize that o.O

Answer 5

so x = 40 and x = 90

Answer 6

dofference is 50

Answer 7

so the answer is 50?

Answer 8

yep..so far so good.

Answer 9

seems right to me

Answer 10

it's all you, shubham.

Answer 11

can you tell me the reasoning for this solution?

Answer 12

x denotes distance..or infact position..
so you get to know that at t=2,,position was as 40 and at 3,,it was at 90
so it travelled 50 metres in between..
which gived you the ans

Answer 13

humm average velocity should be 10 m/s

Answer 14

x denotes distance....then does that mean 10 is speed?

Answer 15

no, x purely informs us about the position of the object in relation to the chosen origin.

Answer 16

why 10 ? o.O

Answer 17

x = 10t^2
v = 20t

between 2 and 3 average speed = (20 x 3 - 20 x2) / 2 = 10 m/s

calculus gives clean answers for these...

Answer 18

so how come the one in between the intervals gives the average velocity?

Answer 19

a particle moves according to the equation x=10t2, where x is in meters and t is in seconds (a) Find the average velocity for the time interval from 2.00 s to 3.00 s

since x is in meters that means it denotes distance/displacement .. and t denotes time .............

this is the case of average instantaneous velocity

Answer 20

We need Feynman to establish a consensus here.

Answer 21

let's not introduce some foreigners @Limitless im xenophibic ;)

Answer 22

well v you obtain by differentiating,,is instantaneous and finding mean of the 2 inst vel. i guess wont be right,,i may be wrong,,please correct me if so..

Answer 23

why is it instantaneous velocity again @mathslover ?

Answer 24

so, if you take the difference between the positions of the object at two different times, we get the displacement of that object.

Answer 25

^right..so how did it become velocity

Answer 26

90 and 40 are positions right?

Answer 27

so wat is the qn............x is distance travelled in that instant r x is the position at time t :/

Answer 28

avg displacement/time=avg velocity

Answer 29

\[x(t)=10t^2\]
\[v_{\text{avg}}=\frac{v_{\text{i}}+v_{\text{f}}}{2}\]

Answer 30

ohh the time is 3 - 2 @rajathsbhat ?

Answer 31

yep.

Answer 32

That was also what I was concluding, Rajath.

Answer 33

Because the interval is \(2\) to \(3\), which leaves room for only \(1\) second.

Answer 34

i see..makes sense..well a follow-up question is to find the average velocity fo tehe time interval from 2.00 to 2.10 s

will it be \[\frac{10(2.10)^2 - 10(2)^2}{2.10-2}?\]

Answer 35

Find the average velocity for the time interval from 2.0 s to 3.0 s.

x(2 s) = 10 (2)2 = 10 (4) = 40 m
x(3 s) = 10 (3)2 = 10 (9) = 90 m

\[x = x_f - x_i = x(3 s) - x(2 s) = 90 m - 40 m = 50 m\]

t = 1.0 s

\[v_avg = x / t = 50 m / 1.0 s = 50 m/s\]

Answer 36

@Limitless' equation for avg velocity is correct as long a the acceleration is constant.

Answer 37

Correct, @rajathsbhat. Aren't we operating under that assumption? I apologize if that is not the case.

Answer 38

the last equation is
\[v_{avg.}=\frac{x}{t}=\frac{50ms^{-1}}{1.0s}=50ms^{-1}\]

Answer 39

yes, we are.

Answer 40

^wait what?!

Answer 41

That's incorrect, @mathslover . That is the velocity, not the average velocity.

Answer 42

where did -1 come from???

Answer 43

and will someone look at my solution for the follow-up question lol

Answer 44

plus, the unit of x is not ms-1

Answer 45

what is this ms-1?! o.O

Answer 46

ms-1 is meter per second...

Answer 47

\[ms ^{-1}\]

Answer 48

oh..a fancy way of writing m/s lo

Answer 49

lol*

Answer 50

lgba, what have you used in ur follow-up sol?

Answer 51

2.10 and 2?

Answer 52

no, the numerator.

Answer 53

Wouldn't
\[v_{avg}=\frac{\frac{x(t_1)}{t_1}+\frac{x(t_2)}{t_2}}{2}\]
be a straightforward formula?

Answer 54

you mean the simplified one?

Answer 55

4.1

Answer 56

@rajathsbhat , is my formula incorrect?

Answer 57

what's 10(2.1)^2 -10(2)^2 supposed to mean?

Answer 58

uhmm substituting t = 2.1 and t = 2

Answer 59

yes @Limitless .Absolutely

Answer 60

that is the final answer

Answer 61

Use my formula, Ig. Let \(t_1=2.1\) and \(t_2=2\).

Answer 62

i wrote the follow-up problem lol..find average velocity for time interval from 2 to 2.10 s

Answer 63

..I know

Answer 64

It still applies.

Answer 65

mehh i dont like formulas @_@

Answer 66

heh i have blue eyes

Answer 67

@lgbasallote see the images that i had uploaded ... that are the final answers..

Answer 68

That's what math is about, lololol.

Answer 69

where did 40 and 60 and 2 come from?

Answer 70

http://assets.openstudy.com/updates/attachments/4fdb077be4b0f2662fd14cdc-mathslover-1339755530702-lgba.png

Answer 71

oh lol it's just a different way of doing the problem?

Answer 72

yes...

Answer 73

and i think that is the easier way to solve that problem

Answer 74

No. @mathslover is calculating velocity, not average velocity.

Answer 75

Unless I'm missing something.

Answer 76

@Limitless i am not calculating the velocity u can see the whole process there is no mistake there in the process .....for calculation of average velocity ...

Answer 77

hmm but i think it would be better to just use this method to be uniform and not get confused...like i said im reluctant to formulas so too much methods gonna clog me up

Answer 78

You used the differential definition of velocity, @mathslover.

Answer 79

anyway i got \[\frac{4.1}{0.1} = 41 m/s\] for the follow up problem

Answer 80

http://en.wikipedia.org/wiki/Velocity#Equation_of_motion

Answer 81

http://www.ux1.eiu.edu/~cfadd/1350/Hmwk/Ch02/Ch2.html

Answer 82

i think there is the answer of limitless reply

Answer 83

http://3mech2011-12.mtlsd.wikispaces.net/file/view/AP+motion+in+1D+Solutions.pdf

see the 3rd question

Answer 84

1) -50 m/s
2) 4.1 m/s

Answer 85

there has been a lot of replies but still none answered me :C can you set aside the arguments later lol

Answer 86

I think whatever @Limitless has said is true. Especially that equation.

Answer 87

can someone show me thesolution for the follow up problem using the 10t^2 method thingy??? just so i know if i did it right

Answer 88

Thank you. I have been eating chips the entire time, so I wasn't able to rigorously check answers.

Answer 89

this thread is long and it's getting laggy so let's be direct to the point lol :P

Answer 90

\[x = 10t^2 \implies v_x = 20t\]
\[v_{avg}=\frac{\int_2^3 20tdt}{\int_2^3 dt} \]idk if it's right but should be

Answer 91

nooo not the calculus way T_T

Answer 92

lgba, there are two methods. This one:
\[v_{avg}=\frac{\frac{x(t_1)}{t_1}-\frac{x(t_2)}{t_2}}{2}\]
Then:
\[
v_{avg}=\frac{1}{b-a}\int_{a}^{b}v(t) dt
\]

Answer 93

why not?