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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(2)=1quad" and "quad v(2)=7
Which of the following expression gives the position of the particle when 
t=0 ?

1+int_(2)^(0)v(t)dt

1+int_(0)^(2)v(t)dt

2+int_(2)^(0)v(t)dt

2+int_(0)^(2)v(t)dt

A particle moves along the x x -axis such that at any time t0 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(2)=1 and v(2)=7 x(2)=1 \quad \text { and } \quad v(2)=7 \newlineWhich of the following expression gives the position of the particle when t=0 t=0 ?\newline1+20v(t)dt 1+\int_{2}^{0} v(t) d t \newline1+02v(t)dt 1+\int_{0}^{2} v(t) d t \newline2+20v(t)dt 2+\int_{2}^{0} v(t) d t \newline2+02v(t)dt 2+\int_{0}^{2} v(t) d t

Full solution

Q. A particle moves along the x x -axis such that at any time t0 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(2)=1 and v(2)=7 x(2)=1 \quad \text { and } \quad v(2)=7 \newlineWhich of the following expression gives the position of the particle when t=0 t=0 ?\newline1+20v(t)dt 1+\int_{2}^{0} v(t) d t \newline1+02v(t)dt 1+\int_{0}^{2} v(t) d t \newline2+20v(t)dt 2+\int_{2}^{0} v(t) d t \newline2+02v(t)dt 2+\int_{0}^{2} v(t) d t
  1. Integrate Velocity Function: To find the position of the particle at time t=0t=0, we need to integrate the velocity function from time t=0t=0 to t=2t=2, since we know the position at t=2t=2 and the velocity at t=2t=2. The integral of the velocity function will give us the change in position from t=0t=0 to t=2t=2.
  2. Calculate Position at t=0t=0: We are given x(2)=1x(2) = 1 and v(2)=7v(2) = 7. To find x(0)x(0), we need to subtract the change in position from t=2t=2 to t=0t=0 from x(2)x(2). This change in position is given by the integral of the velocity function from t=0t=0 to t=2t=2, not from t=2t=2 to t=0t=0. Therefore, we should use the integral with limits from x(2)=1x(2) = 111 to x(2)=1x(2) = 122.
  3. Find Correct Expression: The correct expression for the position of the particle at t=0t=0 is therefore: x(0)=x(2)t=0t=2v(t)dtx(0) = x(2) - \int_{t=0}^{t=2} v(t) \, dt Since x(2)=1x(2) = 1, the expression becomes: x(0)=1t=0t=2v(t)dtx(0) = 1 - \int_{t=0}^{t=2} v(t) \, dt
  4. Choose Correct Option: The correct choice from the given options that represents this expression is: \newline1+t=0t=2v(t)dt1 + \int_{t=0}^{t=2} v(t) \, dt\newlineThis is because the integral of velocity from t=0t=0 to t=2t=2 gives the change in position during this time interval, which when added to the position at t=2t=2, gives the position at t=0t=0.
  5. Final Answer: The correct answer is therefore the second option: 1+t=0t=2v(t)dt1 + \int_{t=0}^{t=2} v(t) \, dt

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