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Consider the following problem: 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? Which expression can we use to solve the problem? Choose 1 answer: (A) int_(6)^(12)a(t)dt (B) a^(')(6)+0.8 (C) a^(')(12) (D) a^(')(12)-a^(')(6)

Consider the following problem:\newlineThe velocity of a pendulum changes at a rate of a(t)=0.9tcos⁡(0.2t) a(t)=0.9^{t} \cos (0.2 t) meters per second squared (where t t is the time in seconds). At time t=6 t=6 , the velocity of the pendulum is 00.88 meters per second. By how much does the velocity change between t=6 t=6 and t=12 t=12 seconds?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) ∫612a(t)dt \int_{6}^{12} a(t) d t \newline(B) a′(6)+0.8 a^{\prime}(6)+0.8 \newline(C) a′(12) a^{\prime}(12) \newline(D) a′(12)−a′(6) a^{\prime}(12)-a^{\prime}(6)

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Q. Consider the following problem:\newlineThe velocity of a pendulum changes at a rate of a(t)=0.9tcos⁡(0.2t) a(t)=0.9^{t} \cos (0.2 t) meters per second squared (where t t is the time in seconds). At time t=6 t=6 , the velocity of the pendulum is 00.88 meters per second. By how much does the velocity change between t=6 t=6 and t=12 t=12 seconds?\newlineWhich expression can we use to solve the problem?\newlineChoose 11 answer:\newline(A) ∫612a(t)dt \int_{6}^{12} a(t) d t \newline(B) a′(6)+0.8 a^{\prime}(6)+0.8 \newline(C) a′(12) a^{\prime}(12) \newline(D) a′(12)−a′(6) a^{\prime}(12)-a^{\prime}(6)
  1. Understand the problem: Understand the problem.\newlineWe are given the acceleration function a(t)=0.9tcos⁡(0.2t)a(t) = 0.9^{t}\cos(0.2 t) and the velocity at t=6t=6 seconds, which is 0.80.8 m/s. We need to find the change in velocity between t=6t=6 and t=12t=12 seconds.
  2. Identify correct expression: Identify the correct expression to use.\newlineTo find the change in velocity, we need to integrate the acceleration function over the time interval from t=6t=6 to t=12t=12 seconds. The integral of acceleration with respect to time gives us the change in velocity.
  3. Choose correct answer: Choose the correct answer from the given options.\newlineThe correct expression to use is the integral of the acceleration function from t=6t=6 to t=12t=12, which is represented by option (A) ∫612a(t) dt\int_{6}^{12}a(t)\,dt.
  4. Calculate change in velocity: Calculate the change in velocity.\newlineWe need to integrate the acceleration function a(t)=0.9tcos⁡(0.2t)a(t) = 0.9^{t}\cos(0.2 t) from t=6t=6 to t=12t=12. This will give us the change in velocity over that time period.
  5. Perform the integration: Perform the integration.\newlineThe actual integration is not provided in the problem, but if we were to carry it out, we would use the integral ∫6120.9tcos⁡(0.2t) dt\int_{6}^{12}0.9^{t}\cos(0.2t)\,dt to find the change in velocity.
  6. Add initial velocity: Add the initial velocity at t=6t=6 to the change in velocity to find the final velocity at t=12t=12. After finding the change in velocity from the integration, we would add it to the initial velocity of 0.8 m/s0.8\,\text{m/s} at t=6t=6 to find the final velocity at t=12t=12.

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