Question

(a). If the position of an ant traveling along a horizontal path at time t is 3t^2+1.
what is the ant's average velocity from t=1 to t=6 ?

Answer 1

\[\Large v_{avg}=\dfrac{v_f+v_i}{2}\]

Answer 2

Now we derive ?
\[3t ^{2}+1\]

Answer 3

Yes

Answer 4

so when we derive 3t^2+1
\[=6t\]

Answer 5

yep, that's right.

Answer 6

so what now?

Answer 7

Did you see the formula I gave you above?

Answer 8

you would need to use that

Answer 9

yup..!

Answer 10

Can you apply that?
vi is velocity at t=1 and vf is velocity at t=6

Answer 11

\[=\frac{ 6+1 }{ 2 }\]
like that?

Answer 12

\(\dfrac{6(6)+6(1)}{2}\)

Answer 13

Because v = 6t

Answer 14

aha yup ..

Answer 15

so the final solution ..

\[=\frac{ 6(6)+6(1) }{ 2 } = \frac{ 42 }{ 2 } = 21\]

Answer 16

Yep

Answer 17

hhhhhhhh

Answer 18

thank you a lot

Answer 19

no problem