Using the equation of motion \(s = ut + \frac{1}{2}at^2\) for the segment \(AB\):

\(55 = 5u + \frac{1}{2}a(5^2)\)

\(55 = 5u + 12.5a\) (Equation 1)

For the segment \(BC\), the final velocity at \(B\) is \(v_B = u + 5a\). Using the equation \(s = ut + \frac{1}{2}at^2\) again:

\(65 = 5v_B + \frac{1}{2}a(5^2)\)

\(65 = 5(u + 5a) + 12.5a\)

\(65 = 5u + 25a + 12.5a\)

\(65 = 5u + 37.5a\) (Equation 2)

Subtract Equation 1 from Equation 2:

\(65 - 55 = (5u + 37.5a) - (5u + 12.5a)\)

\(10 = 25a\)

\(a = 0.4 \text{ m s}^{-2}\)

Substitute \(a = 0.4\) into Equation 1:

\(55 = 5u + 12.5(0.4)\)

\(55 = 5u + 5\)

\(50 = 5u\)

\(u = 10 \text{ m s}^{-1}\)