Consider the following problem: $\newline$ The velocity of a pendulum changes at a rate of $a(t)=0.9^{t} \cos (0.2 t)$ meters per second squared (where $t$ is the time in seconds). At time $t=6$ , the velocity of the pendulum is $0$ . $8$ meters per second. By how much does the velocity change between $t=6$ and $t=12$ seconds? $\newline$ Which expression can we use to solve the problem? $\newline$ Choose $1$ answer: $\newline$ (A) $\int_{6}^{12} a(t) d t$ $\newline$ (B) $a^{\prime}(6)+0.8$ $\newline$ (C) $a^{\prime}(12)$ $\newline$ (D) $a^{\prime}(12)-a^{\prime}(6)$

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Algebra 1

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Q. Consider the following problem:

\newline

The velocity of a pendulum changes at a rate of

a(t)=0.9^{t} \cos (0.2 t)

meters per second squared (where

t

is the time in seconds). At time

t=6

, the velocity of the pendulum is

0

8

meters per second. By how much does the velocity change between

t=6

and

t=12

seconds?

\newline

Which expression can we use to solve the problem?

\newline

Choose

1

answer:

\newline

(A)

\int_{6}^{12} a(t) d t

\newline

(B)

a^{\prime}(6)+0.8

\newline

(C)

a^{\prime}(12)

\newline

(D)

a^{\prime}(12)-a^{\prime}(6)

Understand the problem: Understand the problem. $\newline$ We are given the acceleration function $a(t) = 0.9^{t}\cos(0.2 t)$ and the velocity at $t=6$ seconds, which is $0.8$ m/s. We need to find the change in velocity between $t=6$ and $t=12$ seconds.
Identify correct expression: Identify the correct expression to use. $\newline$ To find the change in velocity, we need to integrate the acceleration function over the time interval from $t=6$ to $t=12$ seconds. The integral of acceleration with respect to time gives us the change in velocity.
Choose correct answer: Choose the correct answer from the given options. $\newline$ The correct expression to use is the integral of the acceleration function from $t=6$ to $t=12$ , which is represented by option (A) $\int_{6}^{12}a(t)\,dt$ .
Calculate change in velocity: Calculate the change in velocity. $\newline$ We need to integrate the acceleration function $a(t) = 0.9^{t}\cos(0.2 t)$ from $t=6$ to $t=12$ . This will give us the change in velocity over that time period.
Perform the integration: Perform the integration. $\newline$ The actual integration is not provided in the problem, but if we were to carry it out, we would use the integral $\int_{6}^{12}0.9^{t}\cos(0.2t)\,dt$ to find the change in velocity.
Add initial velocity: Add the initial velocity at $t=6$ to the change in velocity to find the final velocity at $t=12$ . After finding the change in velocity from the integration, we would add it to the initial velocity of $0.8\,\text{m/s}$ at $t=6$ to find the final velocity at $t=12$ .