We can break the car's motion into three phases:

Phase 1: Uniform Acceleration Initial velocity ($u$) = 0 m/s (starts from rest) Acceleration ($a$) = $2 \text{ m/s}^2$ Time ($t$) = 5 s Final velocity ($v_1$) after this phase: $v_1 = u + at = 0 + (2 \text{ m/s}^2)(5 \text{ s}) = 10 \text{ m/s}$. Distance covered ($s_1$): $s_1 = ut + \frac{1}{2}at^2 = (0)(5) + \frac{1}{2}(2 \text{ m/s}^2)(5 \text{ s})^2 = 0 + 25 \text{ m} = 25 \text{ m}$.

Phase 2: Constant Velocity Velocity ($v$) = $10 \text{ m/s}$ (the final velocity from Phase 1) Time ($t$) = 10 s Distance covered ($s_2$): $s_2 = v \times t = (10 \text{ m/s})(10 \text{ s}) = 100 \text{ m}$.

Phase 3: Uniform Deceleration Initial velocity ($u$) = $10 \text{ m/s}$ (the constant velocity from Phase 2) Final velocity ($v$) = 0 m/s (comes to rest) Acceleration ($a$) = $-4 \text{ m/s}^2$ (deceleration) Distance covered ($s_3$): Using the equation $v^2 = u^2 + 2as$ $0^2 = (10 \text{ m/s})^2 + 2(-4 \text{ m/s}^2)s_3$ $0 = 100 - 8s_3$ $8s_3 = 100 \text{ m}$ $s_3 = \frac{100}{8} \text{ m} = 12.5 \text{ m}$.

Total distance covered by the car: Total Distance = $s_1 + s_2 + s_3 = 25 \text{ m} + 100 \text{ m} + 12.5 \text{ m} = 137.5 \text{ m}$.