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A particle moves along the x-axis such that at any time t >= 0 its position is x(t), its velocity is v(t), and its acceleration is a(t). What is the average acceleration of the particle on the interval 1 <= t <= 10 ? (1)/(9)int_(1)^(10)v(t)dt (a(10)-a(1))/(9) (x(10)-x(1))/(9) (1)/(9)int_(1)^(10)a(t)dt

A particle moves along the x x -axis such that at any time t≥0 t \geq 0 its position is x(t) x(t) , its velocity is v(t) v(t) , and its acceleration is a(t) a(t) .\newlineWhat is the average acceleration of the particle on the interval 1≤t≤10 1 \leq t \leq 10 ?\newline19∫110v(t)dt \frac{1}{9} \int_{1}^{10} v(t) d t \newlinea(10)−a(1)9 \frac{a(10)-a(1)}{9} \newlinex(10)−x(1)9 \frac{x(10)-x(1)}{9} \newline19∫110a(t)dt \frac{1}{9} \int_{1}^{10} a(t) d t

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Q. A particle moves along the x x -axis such that at any time t≥0 t \geq 0 its position is x(t) x(t) , its velocity is v(t) v(t) , and its acceleration is a(t) a(t) .\newlineWhat is the average acceleration of the particle on the interval 1≤t≤10 1 \leq t \leq 10 ?\newline19∫110v(t)dt \frac{1}{9} \int_{1}^{10} v(t) d t \newlinea(10)−a(1)9 \frac{a(10)-a(1)}{9} \newlinex(10)−x(1)9 \frac{x(10)-x(1)}{9} \newline19∫110a(t)dt \frac{1}{9} \int_{1}^{10} a(t) d t
  1. Identify Average Acceleration Formula: To find the average acceleration over a time interval, we need to use the formula for average acceleration, which is the change in velocity Δv\Delta v over the change in time Δt\Delta t. However, the given expressions are not directly related to the average acceleration. We need to identify the correct expression that represents the average acceleration.
  2. Calculate Change in Acceleration: The expression for average acceleration is given by the change in acceleration over the time interval, which is (a(10)−a(1))/(10−1)(a(10) - a(1)) / (10 - 1). This simplifies to (a(10)−a(1))/9(a(10) - a(1)) / 9.
  3. Evaluate Given Expressions: The given expressions are:\newline11. (1)/(9)∫(1)(10)v(t)dt(1)/(9)\int_{(1)}^{(10)}v(t)dt\newline22. (a(10)−a(1))/(9)(a(10)-a(1))/(9)\newline33. (x(10)−x(1))/(9)(x(10)-x(1))/(9)\newline44. (1)/(9)∫(1)(10)a(t)dt(1)/(9)\int_{(1)}^{(10)}a(t)dt\newlineWe need to choose the correct one that represents the average acceleration.
  4. Choose Correct Expression: The correct expression for average acceleration is the second one: (a(10)−a(1))/(9)(a(10)-a(1))/(9). This is because it directly calculates the change in acceleration over the interval from t=1t=1 to t=10t=10 and divides it by the duration of the interval, which is 99 seconds.
  5. Explanation of Incorrect Expressions: The other expressions are incorrect because:\newline11. (1)/(9)∫(1)(10)v(t)dt(1)/(9)\int_{(1)}^{(10)}v(t)dt represents the average velocity, not acceleration.\newline33. (x(10)−x(1))/(9)(x(10)-x(1))/(9) represents the average velocity, assuming constant acceleration, which is not necessarily the case here.\newline44. (1)/(9)∫(1)(10)a(t)dt(1)/(9)\int_{(1)}^{(10)}a(t)dt represents the total change in velocity over the time interval, which is not the same as the average acceleration.

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