a) Write down the equation for the displacement of Car B, \( x_B \):

Car A’s displacement: \( x_A = 4\sqrt{t} \), for \( 0 \leq t \leq 9 \). [1]
Car A reaches position \( x \) at time \( t_A \): \( x = 4\sqrt{t_A} \), so \( t_A = \frac{x^2}{16} \). [1]
Car B starts at \( t = 3 \) and takes half of Car A’s time to reach the same position: \( t_B – 3 = \frac{t_A}{2} = \frac{x^2}{32} \). [1]
Solve: \( x^2 = 32(t – 3) \), so \( x_B = \sqrt{32(t – 3)} = 4\sqrt{2} \sqrt{t – 3} \), for \( t \geq 3 \). [1]
Thus: \( x_B = 4\sqrt{2} \sqrt{t – 3} \), for \( t \geq 3 \). [3]

b) Find the value of \( t \) when Car B catches Car A:

Set \( x_A = x_B \): \( 4\sqrt{t} = 4\sqrt{2} \sqrt{t – 3} \). [1]
Divide by 4: \( \sqrt{t} = \sqrt{2} \sqrt{t – 3} \). [1]
Square both sides: \( t = 2(t – 3) \), so \( t = 2t – 6 \), \( t = 6 \). [1]
Verify: At \( t = 6 \), \( x_A = 4\sqrt{6} \approx 9.798 \), \( x_B = 4\sqrt{2} \cdot \sqrt{3} = 4\sqrt{6} \approx 9.798 \), and \( 6 \in [3, 9] \). [1]
Thus: \( t = 6 \) seconds. [3]

Total: [6 marks]