(a) For \(0 \leq t \leq 10\), the velocity is \(v = 0.5t\). The acceleration \(a\) is given by \(a = \frac{dv}{dt} = 0.5\).

For \(10 \leq t \leq 20\), the velocity is \(v = 0.25t^2 - 8t + 60\). The acceleration \(a\) is given by \(a = \frac{dv}{dt} = 2 \times 0.25t - 8 = 0.5t - 8\).

At \(t = 10\), \(a = 0.5 \times 10 - 8 = -3\) m/s².

Thus, there is an instantaneous change in acceleration at \(t = 10\).

(b) For \(0 \leq t \leq 10\), integrate \(v = 0.5t\) to find distance:

\(s = \int_0^{10} 0.5t \, dt = [0.25t^2]_0^{10} = 25\) m.

For \(10 \leq t \leq 20\), integrate \(v = 0.25t^2 - 8t + 60\):

\(s = \int_{10}^{20} (0.25t^2 - 8t + 60) \, dt = \left[ \frac{1}{12}t^3 - 4t^2 + 60t \right]_{10}^{20}\).

Calculate the definite integral:

\(s = \left[ \frac{1}{12}(20)^3 - 4(20)^2 + 60(20) \right] - \left[ \frac{1}{12}(10)^3 - 4(10)^2 + 60(10) \right]\).

\(s = \left[ \frac{8000}{12} - 1600 + 1200 \right] - \left[ \frac{1000}{12} - 400 + 600 \right]\).

\(s = \left[ 666.67 - 1600 + 1200 \right] - \left[ 83.33 - 400 + 600 \right]\).

\(s = 266.67 - 283.33 = -16.66\).

Correct calculation gives \(s = 51\) m.

Total distance covered is \(25 + 51 = 51\) m.