(i) To find the initial speed, integrate the acceleration to get the velocity function:
\(v(t) = \int a \, dt = \int \left( \frac{3}{160}t^2 - \frac{1}{800}t^3 \right) \, dt = \frac{1}{160}t^3 - \frac{1}{3200}t^4 + C_1\)
Given that the particle comes to rest at B (\(v(20) = 0\)), we have:
\(0 = \frac{1}{160}(20)^3 - \frac{1}{3200}(20)^4 + C_1\)
\(0 = 8000/160 - 160000/3200 + C_1\)
\(C_1 = 0\)
Thus, the initial speed \(v(0) = 0\), confirming the initial speed is zero.
(ii) To find the maximum speed, set the acceleration to zero and solve for \(t\):
\(a = \frac{3}{160}t^2 - \frac{1}{800}t^3 = 0\)
\(t^2(15 - t) = 0\)
\(t = 0\) or \(t = 15\)
The maximum speed occurs at \(t = 15\):
\(v(15) = \frac{1}{160}(15)^3 - \frac{1}{3200}(15)^4 = 5.27 \text{ m/s}\)
(iii) To find the distance AB, integrate the velocity function:
\(s(t) = \int v(t) \, dt = \int \left( \frac{1}{160}t^3 - \frac{1}{3200}t^4 \right) \, dt = \frac{1}{640}t^4 - \frac{1}{16000}t^5 + C_2\)
Using limits from 0 to 20:
\(s(20) - s(0) = \left( \frac{1}{640}(20)^4 - \frac{1}{16000}(20)^5 \right) - 0\)
\(= 250 - 200 = 50 \text{ m}\)
Thus, the distance AB is 50 m.
