# Thread: 293 AP Calculus Exam a(t) and v(t) t=8

1. OK, this can only be done by observation so since we have v(t) I chose e
but the eq should have a minus sign.

here the WIP version of the AP Calculus Exam PDF as created in Overleaf

the goal is to have 365 problems that align basically where students are first year calculus
always appreciate comments since it needs to be a group effort.

2.

3. Originally Posted by karush
OK, this can only be done by observation so since we have v(t) I chose e
but the eq should have a minus sign.
choice (e) works as written only if acceleration remains constant over the indicated interval of time

fyi, the avg value of a function of time over the interval of time $[a,b]$ is $\displaystyle \dfrac{1}{b-a} \int_a^b f(t) \, dt$

which of the other 4 choices matches up?

by f(t) do you mean v(t)?

assuming c with abs enclosure

5. in general, the average value of a function is ...

$\displaystyle \overline{f(x)} = \dfrac{1}{b-a} \int_a^b f(x) \, dx$

in general, if $f$ is any function of time over the time interval $[a,b]$ ...

$\displaystyle \overline{f(t)} = \dfrac{1}{b-a} \int_a^b f(t) \, dt$

so, specifically ...

average acceleration, $\displaystyle \overline{a(t)} = \dfrac{1}{b-a} \int_a^b a(t) \, dt$

average velocity, $\displaystyle \overline{v(t)} = \dfrac{1}{b-a} \int_a^b v(t) \, dt$

average position, $\displaystyle \overline{x(t)} = \dfrac{1}{b-a} \int_a^b x(t) \, dt$

choice (c) is average speed, not velocity.

6.

pl I think B is the answer due to absence of absolute value.

sorry just noticed I never replied to this.

that was a lot of help .... kinda confusing at first.

8. Originally Posted by karush
pl I think B is the answer due to absence of absolute value.

sorry just noticed I never replied to this.

that was a lot of help .... kinda confusing at first.
(B) is correct

$\displaystyle \dfrac{1}{8} \int_0^8 v(t) \, dt = \dfrac{x(8)-x(0)}{8-0} = \dfrac{\Delta x}{\Delta t} = \bar{v}$