A particle moves along the path with velocity function v(t) = t^4 ln(t), t >= 1. Find its position function s(t) = integral_1^t f(x) dx. Evaluate the required integral using integration by parts. You must show all of your steps and how you arrived at your final answer.
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Step 1: Let \( F(x) = x^4 \ln(x) \) and \( S(t) = \int F(x) \, dx \). Show more…
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Let f(t) = F'(t), with F(t) = 4t^2 + 8t, and let a = 2 and b = 3. Write the integral ∫[a to b] f(t) dt and evaluate it using the Fundamental Theorem of Calculus: ∫[a to b] f(t) dt =
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$x=40+12 t-t^{3}$ $\therefore$ Velocity $v=\frac{d x}{d t}=12-3 t^{2}$ When particle comes to rest, $\frac{d x}{d t}=v=0$ $\therefore 12-3 t^{2}=0 \Rightarrow 3 t^{2}=12 \Rightarrow t=2$ seconds Distance travelled by the particle before coming to rest is $\int_{0}^{s} d s=\int_{0}^{2} v d t$ $\therefore s=\int_{0}^{2}\left(12-3 t^{2}\right) d t=\left[12 t-\frac{3 t^{3}}{3}\right]_{0}^{2}$ $=12 \times 2-8=16 \mathrm{~m}$ Hence, the correct answer is option (a).
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