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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). You are given: x(7)=1quad" and "quad v(7)=8 Which of the following expression gives the position of the particle when t=0 ? 1+int_(7)^(0)v(t)dt 1+int_(0)^(7)v(t)dt 7+int_(0)^(7)v(t)dt 7+int_(7)^(0)v(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) . You are given:\newlinex(7)=1 and v(7)=8 x(7)=1 \quad \text { and } \quad v(7)=8 \newlineWhich of the following expression gives the position of the particle when t=0 t=0 ?\newline1+∫70v(t)dt 1+\int_{7}^{0} v(t) d t \newline1+∫07v(t)dt 1+\int_{0}^{7} v(t) d t \newline7+∫07v(t)dt 7+\int_{0}^{7} v(t) d t \newline7+∫70v(t)dt 7+\int_{7}^{0} v(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) . You are given:\newlinex(7)=1 and v(7)=8 x(7)=1 \quad \text { and } \quad v(7)=8 \newlineWhich of the following expression gives the position of the particle when t=0 t=0 ?\newline1+∫70v(t)dt 1+\int_{7}^{0} v(t) d t \newline1+∫07v(t)dt 1+\int_{0}^{7} v(t) d t \newline7+∫07v(t)dt 7+\int_{0}^{7} v(t) d t \newline7+∫70v(t)dt 7+\int_{7}^{0} v(t) d t
  1. Given Data: We are given the position x(7)=1x(7) = 1 and the velocity v(7)=8v(7) = 8. To find the position at time t=0t=0, we need to use the fundamental theorem of calculus, which relates the position function x(t)x(t) to its derivative, the velocity function v(t)v(t). The position x(t)x(t) at any time tt can be found by integrating the velocity function from a known time to the time tt.
  2. Fundamental Theorem of Calculus: Since we know the position at time t=7t=7, we can express the position at time t=0t=0 as x(0)=x(7)+∫t=7t=0v(t) dtx(0) = x(7) + \int_{t=7}^{t=0} v(t) \, dt. This is because the integral of the velocity function over an interval gives the change in position over that interval.
  3. Expression for Position at t=00: The correct integral to express the change in position from time t=7t=7 to time t=0t=0 is ∫t=7t=0v(t) dt\int_{t=7}^{t=0} v(t) \, dt. This integral will give us a negative value since we are integrating backwards (from a larger to a smaller time value). To correct for this, we need to subtract this value from x(7)x(7) to get x(0)x(0).
  4. Correct Expression for Position at t=00: Therefore, the correct expression for the position of the particle at time t=0t=0 is x(0)=x(7)−∫t=7t=0v(t) dtx(0) = x(7) - \int_{t=7}^{t=0} v(t) \, dt. This simplifies to x(0)=1−∫t=7t=0v(t) dtx(0) = 1 - \int_{t=7}^{t=0} v(t) \, dt.
  5. Matching Expression with Given Options: Looking at the given options, the expression that matches our result is 1+∫t=7t=0v(t) dt1 + \int_{t=7}^{t=0} v(t) \, dt. However, we need to be careful with the sign. The integral from t=7t=7 to t=0t=0 will be negative, so we need to subtract it from 11, not add it. Therefore, none of the given options are correct.

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