▶️ Answer/Explanation
Solution
(a)
Set \( v(t) = \ln(t^2 – 4t + 5) – 0.2t = 0 \). Solving numerically, \( t_R \approx 1.426 \).
For \( t = 1 \), \( v(1) = \ln(2) – 0.2 > 0 \), so the particle moves right.
Answer: \( t_R \approx 1.426 \), moving right because \( v(t) > 0 \).
(b)
\( a(t) = v'(t) = \frac{2t – 4}{t^2 – 4t + 5} – 0.2 \). At \( t = 1.5 \):
\( \frac{2(1.5) – 4}{1.5^2 – 4(1.5) + 5} – 0.2 = \frac{-1}{1.25} – 0.2 = -0.8 – 0.2 = -1 \).
\( v(1.5) \approx -0.077 < 0 \), \( a(1.5) = -1 < 0 \), same sign, so speed increases.
Answer: Acceleration = -1, speed increasing.
(c)
\( x(4) = x(1) + \int_1^4 v(t) \, dt = -3 + \int_1^4 (\ln(t^2 – 4t + 5) – 0.2t) \, dt \approx -3 + 0.197 = -2.803 \).
Answer: \( x(4) \approx -2.803 \).
(d)
Total distance = \( \int_1^4 |v(t)| \, dt \). With \( v(t) = 0 \) at \( t \approx 1.426 \),
\( \int_1^{1.426} v(t) \, dt + \int_{1.426}^4 -v(t) \, dt \approx 0.958 \).
Answer: 0.958.