Question

The maximum possible acceleration of a train moving on a straight track is 10 m/s² and maximum possible retardation is 5 m/s². If maximum achievable speed of train is 10 m/s then minimum time (in s) in which train can complete a journey of 135 m starting from rest and ending at rest, is :

Answer 1

15

Answer 2

To find the minimum time for the train to complete the journey, we need to utilize the maximum possible acceleration and retardation, and if the distance allows, reach the maximum achievable speed.

The journey can be divided into three phases for minimum time:

1. Acceleration Phase: The train accelerates from rest ($u=0$) with maximum acceleration ($a_{max} = 10 \text{ m/s}^2$) until it reaches its maximum speed ($v_{max} = 10 \text{ m/s}$).
2. Constant Velocity Phase: The train moves at its maximum speed ($v_{max} = 10 \text{ m/s}$). This phase occurs only if the total distance is large enough.
3. Retardation Phase: The train decelerates with maximum retardation ($r_{max} = 5 \text{ m/s}^2$) from $v_{max}$ until it comes to rest ($v=0$).

Let's calculate the time and distance for the acceleration and retardation phases:

Phase 1: Acceleration

• Initial velocity ($u_1$) = 0 m/s
• Final velocity ($v_1$) = $v_{max}$ = 10 m/s
• Acceleration ($a_1$) = $a_{max}$ = 10 m/s²

Using the kinematic equation $v = u + at$:

$10 = 0 + 10 \times t_1$

$t_1 = 1 \text{ s}$

Using the kinematic equation $v^2 = u^2 + 2aS$:

$10^2 = 0^2 + 2 \times 10 \times S_1$

$100 = 20 S_1$

$S_1 = 5 \text{ m}$

Phase 3: Retardation

• Initial velocity ($u_3$) = $v_{max}$ = 10 m/s
• Final velocity ($v_3$) = 0 m/s
• Retardation ($a_3$) = $r_{max}$ = 5 m/s² (we use positive value for calculation, just remember it's deceleration)

Using the kinematic equation $v = u - at$:

$0 = 10 - 5 \times t_3$

$5 t_3 = 10$

$t_3 = 2 \text{ s}$

Using the kinematic equation $v^2 = u^2 - 2aS$:

$0^2 = 10^2 - 2 \times 5 \times S_3$

$0 = 100 - 10 S_3$

$10 S_3 = 100$

$S_3 = 10 \text{ m}$

Check if the train reaches maximum speed:

The total distance covered during acceleration and retardation phases is $S_{accel+retard} = S_1 + S_3 = 5 \text{ m} + 10 \text{ m} = 15 \text{ m}$.

The total journey distance is $S_{total} = 135 \text{ m}$.

Since $S_{accel+retard} (15 \text{ m}) < S_{total} (135 \text{ m})$, the train does reach its maximum speed and travels at it for some duration.

Phase 2: Constant Velocity

The distance covered at constant maximum speed ($S_2$) is the total distance minus the distances covered during acceleration and retardation:

$S_2 = S_{total} - S_1 - S_3 = 135 \text{ m} - 5 \text{ m} - 10 \text{ m} = 120 \text{ m}$.

The time taken for this phase ($t_2$) is:

$t_2 = S_2 / v_{max} = 120 \text{ m} / 10 \text{ m/s} = 12 \text{ s}$.

Total Minimum Time

The total minimum time ($T_{min}$) for the journey is the sum of the times for all three phases:

$T_{min} = t_1 + t_2 + t_3 = 1 \text{ s} + 12 \text{ s} + 2 \text{ s} = 15 \text{ s}$.

The velocity-time graph for this motion would be a trapezium, starting from 0, rising to 10 m/s, staying at 10 m/s, and then falling to 0. The area under this graph would be 135 m.

The final answer is $\boxed{\text{15}}$.

Explanation of the solution:

To minimize time, the train accelerates at its maximum possible rate (10 m/s²) to reach its maximum speed (10 m/s), travels at that maximum speed, and then decelerates at its maximum possible rate (5 m/s²) to come to rest.

1. Acceleration phase: Time taken ($t_1$) to reach 10 m/s from rest at 10 m/s² is 1 s. Distance covered ($S_1$) is 5 m.
2. Retardation phase: Time taken ($t_3$) to stop from 10 m/s at 5 m/s² is 2 s. Distance covered ($S_3$) is 10 m.
3. Check for constant speed phase: The sum of distances for acceleration and retardation is $5 \text{ m} + 10 \text{ m} = 15 \text{ m}$. Since the total distance is 135 m (which is greater than 15 m), the train will indeed travel at maximum speed for some duration.
4. Constant speed phase: The remaining distance is $135 \text{ m} - 15 \text{ m} = 120 \text{ m}$. Time taken ($t_2$) to cover 120 m at 10 m/s is 12 s.
5. Total time: Summing the times from all phases: $1 \text{ s} + 12 \text{ s} + 2 \text{ s} = 15 \text{ s}$.