A car can be accelerated with maximum acceleration of 4m/s^2 and it brakes can produce a maximum retardation of 2 m/s^2.

Question

A car can be accelerated with maximum acceleration of 4m/s^2 and it brakes can produce a maximum retardation of 2 m/s^2. The minimum time to cover a distance of 1200 m starting from rest and finally coming to rest, is

Answer 1

Using v_f² = v_i² + 2ax

For Section 1: v² = (2)(4)(x) = 8x

For Section 2: 0² = v² + (2)(-2)(1200 - x) or v² = 4800 - 4x

Setting those two equations equal gives: 8x = 4800 - 4x or 12x = 4800 or x = 400

For Section 1, using x_f = x_i + v_it + 1/2at₁²: 400 = 1/2(4)(t₁²) or t₁ = √200

Using v_f = v_i + at: v_f = 4√200

For Section 2: using v_f = v_i + at₂: 0 = 4√200 + (-2)(t₂) or t₂ = 2√200

Total time = t₁ + t₂ = √200 + 2√200 = 3√200 = 42.43 s