1.4.293 AP Calculus Exam a(t) and v(t) t=8

[karush]

View attachment 9508
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

https://documentcloud.adobe.com/link/track?uri=urn:aaid:scds:US:053a75d8-ca5b-4447-bd65-4e580f0de793

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.

 

[skeeter]

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

which of the other 4 choices matches up?

 

[karush]

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

assuming c with abs enclosure

 

[skeeter]

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.

 

[skeeter]

AP Calculus resource ...

https://apcentral.collegeboard.org/pdf/ap-curricmodcalculusmotion.pdf?course=ap-calculus-bc

 

[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.

 

[skeeter]

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

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