The position of the particle moving along the x-axis when t = 4 is approximately 1.352. (Correct choice: C)
By mechanical physics and integral calculus, the function position (x(t)) represents the indefinite integral of the function velocity (v(t)). That is to say:
[tex]x(t) = \int {v(t)} \, dt[/tex] Â (1)
First, we find the function position by (1):
[tex]x(t) = 4\int {\frac{dt}{t^{3}+1} } + C[/tex]
[tex]x(t) = \frac{2}{3}\cdot \left[-\ln (t^{2}-t+1)+2\cdot \ln (t+1)+2\sqrt{3}\cdot \tan^{-1} \left(\frac{2\cdot t - 1}{\sqrt{3}} \right)\right] + C[/tex]
Where [tex]C[/tex] is the integration constant.
If we know that [tex]x(2) = 1[/tex], then the integration constant is:
[tex]1 = \frac{2}{3}\cdot \left[-\ln(2^{2}-2+1)+2\cdot \ln(2+1)+2\sqrt{3}\cdot \tan^{-1}\left(\frac{2\cdot 2 - 1}{\sqrt{3}} \right)\right] +C[/tex]
[tex]1 = 3.151 + C[/tex]
[tex]C = -2.151[/tex]
Hence, the function position of the particle is:
[tex]x(t) = \frac{2}{3}\cdot \left[-\ln (t^{2}-t+1)+2\cdot \ln (t+1)+2\sqrt{3}\cdot \tan^{-1} \left(\frac{2\cdot t - 1}{\sqrt{3}} \right)\right] -2.151[/tex]
If we know that [tex]t = 4[/tex], then the position of the particle is:
[tex]x(4) = \frac{2}{3}\cdot \left[-\ln (4^{2}-4+1)+2\cdot \ln (4+1)+2\sqrt{3}\cdot \tan^{-1} \left(\frac{2\cdot 4 - 1}{\sqrt{3}} \right)\right] -2.151[/tex]
[tex]x(4) \approx 1.352[/tex]
The position of the particle moving along the x-axis when t = 4 is approximately 1.352. (Correct choice: C) [tex]\blacksquare[/tex]
The statement is incomplete, poorly formated and full of typing mistakes. Correct statement is described herein:
A particle moves along the x-axis. The velocity of the particle at time t is given by [tex]v(t) = \frac{4}{t^{3}+1}[/tex]. If the position of the particle is x = 1 when t = 2, what is the position of the particle when t = 4?
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