1. Analyze the Motion Constraints

The motion of the drive consists of three main stages: acceleration to maximum speed, constant velocity (cruising), and deceleration to stop. To find the shortest transit time, the drive must accelerate to its maximum possible speed as quickly as possible (within safety limits), cruise, and then decelerate.

Given parameters:

• $D = 2.0 \text{ km} = 2000 \text{ m}$
• $a_{max} = 4.0 \text{ m/s}^2$
• $j_{max} = 1.0 \text{ m/s}^3$
• $v_{max} = 144 \text{ km/h} = 40 \text{ m/s}$

2. Analysis of the Acceleration Phase

We first calculate the time and distance required to reach the maximum speed of $40 \text{ m/s}$. The acceleration cannot jump instantly to $4.0 \text{ m/s}^2$ due to the jerk limit. It must ramp up linearly.

Step A: Ramping up acceleration (Jerk Limited) In this interval, jerk is constant at $+1 \text{ m/s}^3$.
Time to reach max acceleration ($t_1$): $$t_1 = \frac{a_{max}}{j_{max}} = \frac{4.0}{1.0} = 4 \text{ s}$$ Velocity gained ($v_1$): $$v_1 = \int_0^4 a(t) dt = \int_0^4 (1 \cdot t) dt = \left[ \frac{t^2}{2} \right]_0^4 = 8 \text{ m/s}$$ Distance covered ($S_1$): $$S_1 = \int_0^4 v(t) dt = \int_0^4 \frac{t^2}{2} dt = \left[ \frac{t^3}{6} \right]_0^4 = \frac{64}{6} = \frac{32}{3} \text{ m}$$

Step B: Constant Acceleration By symmetry, the “ramp down” of acceleration (Step C) will also take $4 \text{ s}$ and contribute another $8 \text{ m/s}$ to the velocity change. $$\Delta v_{ramps} = 8 \text{ m/s} + 8 \text{ m/s} = 16 \text{ m/s}$$ Velocity needed during the constant acceleration phase to reach $v_{max} = 40 \text{ m/s}$: $$\Delta v_{const} = 40 – 16 = 24 \text{ m/s}$$ Time required for this phase ($t_2$): $$t_2 = \frac{\Delta v_{const}}{a_{max}} = \frac{24}{4} = 6 \text{ s}$$ Distance covered ($S_2$): Since initial velocity for this phase is $8 \text{ m/s}$: $$S_2 = v_{start}t_2 + \frac{1}{2}a_{max}t_2^2 = 8(6) + \frac{1}{2}(4)(6^2) = 48 + 72 = 120 \text{ m}$$

Step C: Ramping down acceleration In this phase, acceleration decreases linearly from $4 \text{ m/s}^2$ to $0$. The jerk is $-1 \text{ m/s}^3$.
Initial velocity $u_3$ (at start of this phase) is $8 + 24 = 32 \text{ m/s}$.
Duration $t_3 = 4 \text{ s}$.
Acceleration function: $a(t) = 4 – t$ (for $t \in [0,4]$).
Velocity function: $$v(t) = u_3 + \int_0^t a(\tau) d\tau = 32 + \left[ 4\tau – \frac{\tau^2}{2} \right]_0^t = 32 + 4t – \frac{t^2}{2}$$ Displacement $S_3$ involves integrating the velocity function: $$S_3 = \int_0^4 v(t) dt = \int_0^4 \left( 32 + 4t – \frac{t^2}{2} \right) dt$$ $$S_3 = \left[ 32t + 2t^2 – \frac{t^3}{6} \right]_0^4$$ $$S_3 = 32(4) + 2(16) – \frac{64}{6} = 128 + 32 – \frac{32}{3}$$ $$S_3 = 160 – 10.67 = 149.33 \text{ m} \quad \left(\text{exact: } \frac{448}{3} \text{ m}\right)$$

Total Acceleration Phase:
Total Time $T_{acc} = 4 + 6 + 4 = 14 \text{ s}$.
Total Distance $S_{acc} = S_1 + S_2 + S_3 = \frac{32}{3} + 120 + \frac{448}{3} = \frac{480}{3} + 120 = 160 + 120 = 280 \text{ m}$.

3. Calculation of Total Time

The total distance is $2000 \text{ m}$.
Distance covered during acceleration: $S_{acc} = 280 \text{ m}$.
Distance covered during deceleration (symmetric to acceleration): $S_{dec} = 280 \text{ m}$.

Remaining distance for constant velocity cruising: $$S_{cruise} = D_{total} – S_{acc} – S_{dec} = 2000 – 280 – 280 = 1440 \text{ m}$$

Time spent cruising: $$T_{cruise} = \frac{S_{cruise}}{v_{max}} = \frac{1440}{40} = 36 \text{ s}$$

Total transit time: $$T_{total} = T_{acc} + T_{cruise} + T_{dec}$$ $$T_{total} = 14 + 36 + 14 = 64 \text{ s}$$