To solve the 18th question about the two electric trains, we follow these steps:

Given:

• Train A:
  • Acceleration ($a_A$) = 0.2 m/s²
  • Final speed ($v_A$) = 45 km/h = 12.5 m/s
• Train B:
  • Starts 1 minute (60 seconds) later
  • Acceleration ($a_B$) = 0.4 m/s²
  • Final speed ($v_B$) = 72 km/h = 20 m/s

Find: When Train B overtakes Train A.

Steps:

1. Time for Train A to reach its maximum speed: $v_A = u_A + a_A \times t_A$ $12.5 = 0 + 0.2 \times t_A \implies t_A = 62.5 \text{ seconds}$

2. Distance covered by Train A during acceleration: $s_A = u_A \times t_A + \frac{1}{2} a_A \times t_A^2$ $s_A = 0 + \frac{1}{2} \times 0.2 \times (62.5)^2 \implies s_A = 390.625 \text{ meters}$

3. Time for Train B to reach its maximum speed: $v_B = u_B + a_B \times t_B$ $20 = 0 + 0.4 \times t_B \implies t_B = 50 \text{ seconds}$

4. Distance covered by Train B during acceleration: $s_B = u_B \times t_B + \frac{1}{2} a_B \times t_B^2$ $s_B = 0 + \frac{1}{2} \times 0.4 \times (50)^2 \implies s_B = 500 \text{ meters}$

5. Relative motion after each train reaches its maximum speed:

   • Train A has constant speed after 62.5 seconds
   • Train B becomes constant speed after 50 seconds

Train B starts 60 seconds later, so:

• Total time for Train A until it is overtaken = t
• Total time for Train B to overtake = t - 60

Set the equations for the point of overtaking:

For Train A: Total distance covered = $390.625 + 12.5 \times (t - 62.5)$

For Train B: After acceleration, distance covered = $500 + 20 \times (t - 110)$

Set the two distances equal: $390.625 + 12.5 \times (t - 62.5) = 500 + 20 \times (t - 110)$

Solve for $t$: $390.625 + 12.5t - 781.25 = 500 + 20t - 2200$ $12.5t - 20t = 500 - 2200 + 390.625 - 781.25$ $-7.5t = -2090.625$ $t = 278.75 \text{ seconds}$

Conclusion: Train B will overtake Train A 278.75 seconds after Train A starts moving.