– Let east be the positive \( x \)-direction.
– **Red car** starts at \( x_R(0) = 0 \) with \( v_{R0} = 0 \) and acceleration \( a_R = +3.5 \, \text{m/s}^2 \).
– **Blue car** starts at \( x_B(0) = 600 \, \text{m} \) with \( v_{B0} = -15 \, \text{m/s} \) (since it’s moving west) and acceleration \( a_B = -1.2 \, \text{m/s}^2 \).

– **Red car position:**
\( x_R(t) = v_{R0} t + \tfrac{1}{2} a_R t^2 \)
\( x_R(t) = 0 + \tfrac{1}{2} (3.5) t^2 = 1.75 t^2 \)

– **Blue car position:**
\( x_B(t) = x_B(0) + v_{B0} t + \tfrac{1}{2} a_B t^2 \)
\( x_B(t) = 600 + (-15) t + \tfrac{1}{2} (-1.2) t^2 \)
\( x_B(t) = 600 – 15 t – 0.6 t^2 \)

\( x_R(t) = x_B(t) \)
\( 1.75 t^2 = 600 – 15 t – 0.6 t^2 \)

\( 1.75 t^2 + 0.6 t^2 + 15 t – 600 = 0 \)
\( 2.35 t^2 + 15 t – 600 = 0 \)

Quadratic equation: \( a t^2 + b t + c = 0 \)
Coefficients: \( a = 2.35 \), \( b = 15 \), \( c = -600 \)

Discriminant:
\( D = b^2 – 4 a c = (15)^2 – 4 (2.35)(-600) = 225 + 5640 = 5865 \)

\( t = \dfrac{ -b \pm \sqrt{D} }{ 2 a } = \dfrac{ -15 \pm \sqrt{5865} }{ 4.7 } \)

\( \sqrt{5865} \approx 76.609 \)
\( t = \dfrac{ -15 + 76.609 }{ 4.7 } \approx \dfrac{61.609}{4.7} \approx 13.11 \, \text{s} \)

(Negative root yields negative time, which is not physical in this context.)

– Using red car’s position:
\( x_R = 1.75 t^2 = 1.75 \times (13.11)^2 \approx 1.75 \times 171.8521 \approx 300.74 \, \text{m} \)

– Using blue car’s position:
\( x_B = 600 – 15 t – 0.6 t^2 \)
\( x_B = 600 – 15 \times 13.11 – 0.6 \times (13.11)^2 \)
\( x_B \approx 600 – 196.65 – 103.11 \approx 300.24 \, \text{m} \)

– Slight difference due to rounding; positions agree within calculation accuracy.

– Red car’s velocity:
\( v_R = v_{R0} + a_R t = 0 + (3.5)(13.11) \approx 45.89 \, \text{m/s} \)

– Blue car’s velocity:
\( v_B = v_{B0} + a_B t = -15 + (-1.2)(13.11) \approx -30.73 \, \text{m/s} \)