Questions: Determine the time (in seconds) required for a train to travel 1 [km] along its rails if the train starts from rest, reaches a maximum speed at some intermediate point, and then stops at the end of the road. The train can accelerate at 1.5 [m /(s^2)] and decelerate at 2[m /(s^2)].

Determine the time (in seconds) required for a train to travel 1 [km] along its rails if the train starts from rest, reaches a maximum speed at some intermediate point, and then stops at the end of the road. The train can accelerate at 1.5 [m /(s^2)] and decelerate at 2[m /(s^2)].

Transcript text: Determine the time (in seconds) required for a train to travel 1 [ km ] along its rails if the train starts from rest, reaches a maximum speed at some intermediate point, and then stops at the end of the road. The train can accelerate at 1.5 $\left[\mathrm{m} /\left(\mathrm{s}^{\wedge} 2\right)\right]$ and decelerate at $2\left[\mathrm{~m} /\left(\mathrm{s}^{\wedge} 2\right)\right]$.

Solution

Solution Steps

Step 1: Define the Problem and Given Data

Distance to travel: $ d = 1 \, \text{km} = 1000 \, \text{m} $
Acceleration: $ a_1 = 1.5 \, \text{m/s}^2 $
Deceleration: $ a_2 = 2 \, \text{m/s}^2 $
Initial speed: $ v_0 = 0 \, \text{m/s} $
Final speed: $ v_f = 0 \, \text{m/s} $

Step 2: Determine Maximum Speed

Let $ v_m $ be the maximum speed reached.
Using the kinematic equation for acceleration: $ v_m^2 = v_0^2 + 2a_1d_1 $
For deceleration: $ v_f^2 = v_m^2 - 2a_2d_2 $
Since $ v_f = 0 $, $ v_m^2 = 2a_2d_2 $
Total distance: $ d = d_1 + d_2 $

Step 3: Solve for Distances $ d_1 $ and $ d_2 $

From $ v_m^2 = 2a_1d_1 $ and $ v_m^2 = 2a_2d_2 $, equate the two: \[ 2a_1d_1 = 2a_2d_2 \implies a_1d_1 = a_2d_2 \]
Using $ d = d_1 + d_2 $: \[ d_1 = \frac{a_2}{a_1 + a_2} \cdot d \quad \text{and} \quad d_2 = \frac{a_1}{a_1 + a_2} \cdot d \]

Step 4: Calculate $ d_1 $ and $ d_2 $

$ d_1 = \frac{2}{1.5 + 2} \cdot 1000 \, \text{m} = \frac{2}{3.5} \cdot 1000 \, \text{m} \approx 571.43 \, \text{m} $
$ d_2 = \frac{1.5}{1.5 + 2} \cdot 1000 \, \text{m} = \frac{1.5}{3.5} \cdot 1000 \, \text{m} \approx 428.57 \, \text{m} $

Step 5: Calculate Maximum Speed $ v_m $

Using $ v_m^2 = 2a_1d_1 $: \[ v_m = \sqrt{2 \cdot 1.5 \cdot 571.43} \approx \sqrt{1714.29} \approx 41.41 \, \text{m/s} \]

Step 6: Calculate Time for Acceleration and Deceleration

Time to accelerate to $ v_m $: \[ t_1 = \frac{v_m - v_0}{a_1} = \frac{41.41 - 0}{1.5} \approx 27.61 \, \text{s} \]
Time to decelerate from $ v_m $: \[ t_2 = \frac{v_m - v_f}{a_2} = \frac{41.41 - 0}{2} \approx 20.71 \, \text{s} \]

Step 7: Calculate Total Time