If the greatest admissible acceleration or retardation of a train be 3 feet/sec2.find the least time from one station to another at a distance of 10m the maximum speed being 60 miles per hour:

To determine the least time for a train to travel between two stations, given the constraints of acceleration, deceleration, and maximum speed, we can break down the problem into manageable steps. Let's start by converting the maximum speed into a more usable unit and then apply the equations of motion.

Step 1: Convert Maximum Speed

The maximum speed of the train is given as 60 miles per hour. To work with this in our calculations, we need to convert it to feet per second.

• 1 mile = 5280 feet
• 1 hour = 3600 seconds

Thus, the conversion is:

Maximum Speed (V) = 60 miles/hour × (5280 feet/mile) / (3600 seconds/hour) = 88 feet/second.

Step 2: Analyze the Motion

We need to consider the train's motion in three phases: acceleration, constant speed, and deceleration. The total distance covered is 10 meters, which we will convert to feet for consistency:

Distance (D) = 10 meters × 3.281 feet/meter = 32.81 feet.

Step 3: Determine Acceleration and Deceleration Phases

The maximum admissible acceleration and retardation (deceleration) of the train is 3 feet/sec². We will use this to calculate the time taken during the acceleration and deceleration phases.

Acceleration Phase

Using the formula for distance during acceleration:

D_a = (1/2) * a * t_a²

Where:

• D_a = distance during acceleration
• a = acceleration (3 feet/sec²)
• t_a = time taken to accelerate

We also know that the final speed at the end of the acceleration phase can be expressed as:

V = a * t_a

Substituting for t_a gives:

t_a = V/a = 88 feet/sec / 3 feet/sec² = 29.33 seconds.

Now, substituting t_a back into the distance equation:

D_a = (1/2) * 3 * (29.33)² = 43.33 feet.

Deceleration Phase

Similarly, during deceleration, the distance covered can be calculated using the same formula. The time taken to decelerate from maximum speed to rest is the same as the time taken to accelerate:

t_d = t_a = 29.33 seconds.

Thus, the distance during deceleration is also:

D_d = (1/2) * 3 * (29.33)² = 43.33 feet.

Step 4: Total Distance and Time Calculation

Now, we can find the total distance covered during acceleration and deceleration:

D_total = D_a + D_d = 43.33 + 43.33 = 86.66 feet.

Since the total distance (32.81 feet) is less than the distance covered during acceleration and deceleration, the train will not reach its maximum speed. Therefore, we need to recalculate the time based on the total distance.

Revised Time Calculation

Using the total distance of 32.81 feet, we can set up the equation:

D = (1/2) * a * t²

Solving for t gives:

t = sqrt((2 * D) / a) = sqrt((2 * 32.81) / 3) = sqrt(21.87) ≈ 4.67 seconds.

Final Time Calculation

Since the train accelerates for 4.67 seconds and does not reach maximum speed, the total time taken to travel the distance of 10 meters (32.81 feet) is approximately:

Total Time ≈ 4.67 seconds.

In summary, the least time for the train to travel from one station to another, under the given conditions, is approximately 4.67 seconds. This analysis illustrates how acceleration and distance interact in motion, and how to apply kinematic equations effectively.