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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(5)=7quad" and "quad v(5)=4
Which of the following expression gives the velocity of the particle when 
t=0 ?

4+int_(0)^(5)x(t)dt

4+int_(5)^(0)x(t)dt

4+int_(5)^(0)a(t)dt

4+int_(0)^(5)a(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(5)=7 and v(5)=4 x(5)=7 \quad \text { and } \quad v(5)=4 \newlineWhich of the following expression gives the velocity of the particle when t=0 t=0 ?\newline4+05x(t)dt 4+\int_{0}^{5} x(t) d t \newline4+50x(t)dt 4+\int_{5}^{0} x(t) d t \newline4+50a(t)dt 4+\int_{5}^{0} a(t) d t \newline4+05a(t)dt 4+\int_{0}^{5} a(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(5)=7 and v(5)=4 x(5)=7 \quad \text { and } \quad v(5)=4 \newlineWhich of the following expression gives the velocity of the particle when t=0 t=0 ?\newline4+05x(t)dt 4+\int_{0}^{5} x(t) d t \newline4+50x(t)dt 4+\int_{5}^{0} x(t) d t \newline4+50a(t)dt 4+\int_{5}^{0} a(t) d t \newline4+05a(t)dt 4+\int_{0}^{5} a(t) d t
  1. Given Data: We are given the position and velocity of the particle at time t=5t=5, which are x(5)=7x(5)=7 and v(5)=4v(5)=4, respectively. To find the velocity at time t=0t=0, we need to use the relationship between velocity and acceleration. The velocity function v(t)v(t) is the antiderivative of the acceleration function a(t)a(t). Therefore, we can express the change in velocity from time t=0t=0 to t=5t=5 as the integral of the acceleration function over that interval.
  2. Relationship Between Velocity and Acceleration: The correct expression for the velocity at time t=0t=0 using the Fundamental Theorem of Calculus is v(0)=v(5)50a(t)dtv(0) = v(5) - \int_{5}^{0} a(t) \, dt. This is because we are looking for the velocity at an earlier time (t=0t=0), so we subtract the integral of the acceleration from the known velocity at t=5t=5.
  3. Application of Fundamental Theorem of Calculus: The given options are:\newlineA) 4+05x(t)dt4 + \int_{0}^{5} x(t) \, dt\newlineB) 4+50x(t)dt4 + \int_{5}^{0} x(t) \, dt\newlineC) 4+50a(t)dt4 + \int_{5}^{0} a(t) \, dt\newlineD) 4+05a(t)dt4 + \int_{0}^{5} a(t) \, dt\newlineWe can immediately eliminate options A and B because they involve the integral of the position function x(t)x(t), not the acceleration function a(t)a(t). We need to use the acceleration function to find the change in velocity.
  4. Elimination of Incorrect Options: Between options C and D, option C has the correct limits of integration for finding the velocity at an earlier time t=0t=0 from a known time t=5t=5. Therefore, the correct expression for the velocity at time t=0t=0 is:\newlinev(0)=v(5)50a(t)dtv(0) = v(5) - \int_{5}^{0} a(t) \, dt\newlineThis simplifies to:\newlinev(0)=450a(t)dtv(0) = 4 - \int_{5}^{0} a(t) \, dt
  5. Selection of Correct Option: However, we must be careful with the limits of integration. The integral from 55 to 00 of a(t)a(t) dt is the negative of the integral from 00 to 55 of a(t)a(t) dt. Therefore, we can rewrite the expression as:\newlinev(0)=4+05a(t)dtv(0) = 4 + \int_{0}^{5} a(t) \, dt\newlineThis matches option D, which is the correct expression for the velocity of the particle at time t=0t=0.

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