For the car moving up the hill, apply Newton's second law:

\(1200 - 350 - 1250 \times 10 \times 0.05 = 1250a\)

\(a = \frac{225}{1250} = 0.18 \text{ m/s}^2\)

Using the equation \(v^2 = u^2 + 2as\) with \(u = 0\), \(s = 100\):

\(v^2 = 0 + 2 \times 0.18 \times 100\)

\(v = 6 \text{ m/s}\)

For the car moving down the hill, apply Newton's second law:

\(1200 - 350 + 1250 \times 10 \times 0.05 = 1250a\)

\(a = \frac{1475}{1250} = 1.18 \text{ m/s}^2\)

Using the equation \(v^2 = u^2 + 2as\) with \(u = 0\), \(s = 100\):

\(v^2 = 0 + 2 \times 1.18 \times 100\)

\(v = 15.4 \text{ m/s}\)