Question

Find the average velocity vector during the time interval between t=2 and t=3

v(t)= 2i - 6tj

Answer 1

I've substituted the values 2 and 3 into the velocity vector and got the answers;

V(2) = 2i - 12j
v(3) = 21 -18j

Answer 2

then velocity average = [v(3) -V(2)]/2

Answer 3

but my answers doesn't match the back of the book;(

Answer 4

the average velocity from t = a to t = b is:\[\frac{v(b)-v(a)}{b-a}\]is this the formula you are using?

Answer 5

oh i thought you just divide v(b)-v(a) by 2

Answer 6

you need to divide it by the amount of time that elapsed during your interval, which is 3-2=1 second. i might be wrong, im not much of a physics person.

Answer 7

Guys, you can't average at the beginning and the end because velocity is not increasing linearly. You need to integrate, evaluate the definite integral and then divide by the total time.

Answer 8

my answer was -15j but the back of the book says 2i-15j

Answer 9

pls giv medal to me

Answer 10

ah i see, my bad >.<

Answer 11

oh wait, sorry, it does say t and not t^2, so yeah, averaging the endpoints works. my bad

Answer 12

lol mi sorta confused

Answer 13

is my method correct?

Answer 14

yes, you can evaluate the vector at each endpoint, add them together, and then divide by the time difference

Answer 15

if the answer is 2i-15j, it seems you need to integrate, in which you get 2ti-3t^2j, then plug in the values t = 2 and 3. This gives

4i-12j
6i-27j

then subtract the bottom from the first and divide by 3-2=1. that gives

2i-15j

Answer 16

Formulas

\[Average Velocity= \frac{Final Position- Initial Position}{Time Interval}\]
So

\[Postition =\int Velocity \;\;dt\]
\[Position=\int (2)i-(6t)j \;\;dt \;\;= (2t)i-(3t^{3})j\]

So,
\[Average \; velocity= \frac{\int_{2}^{3} [(2)i-(6t)j]dt}{3-2} \;\;dt \]
\[Average \; velocity= (2t)i-(3t^{3})j \|^{3}_{2}\]
\[Average \; velocity=(2)i-(3(3^2-2^2))j= 2i-15j\]

and Bingo.

Answer 17

a much better explanation lol

Answer 18

wow thanks heaps guys:)