Question

A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1 m/s^2 for 20s. It then proceeds at constant speed for 1100 m before slowing down at 2.2 m/s^2 until it stops at the station. What is the distance between the stations? How much time does it take the train to go between the stations?

Answer 1

Answer of question #79950 -Physics- Mechanics - Relativity

A light-rail train going from one station to the next on a straight section of track accelerates from rest at $1.1 \, \text{m/s}^2$ for 20s. It then proceeds at constant speed for $1100 \, \text{m}$ before slowing down at $2.2 \, \text{m/s}^2$ until it stops at the station. What is the distance between the stations? How much time does it take the train to go between the stations?

Input Data:

Acceleration:

$$a_1 = 1.1 \frac{m}{s^2}$$

Acceleration time: $t_1 = 20 \, \text{s}$

Distance at constant speed: $S_2 = 1100 \, \text{m}$

Slow down to a stop: $a_2 = 2.2 \frac{m}{s^2}$

Solution:

Constant speed, dialed by train:

$$V_1 = a_1 t_1 = 1.1 * 20 = 22 \frac{m}{s}$$

The distance traveled during acceleration:

$$S_1 = \frac{a_1 t_1^2}{2} = 1.1 * \frac{400}{2} = 220 \, \text{m}$$

Travel time at a constant speed:

$$t_2 = \frac{S_2}{V_1} = \frac{1100}{22} = 50 \, \text{s}$$

Braking distance:

$$S_3 = \frac{a_2 t_3^2}{2} = \frac{2.2 * 10^2}{2} = 110 \, \text{m}$$

Distance from station to station:

$$S = S_1 + S_2 + S_3 = 220 + 1100 + 110 = 1430 \, \text{m}$$

Total travel time from station to station:

$$t = t_1 + t_2 + t_3 = 20 + 50 + 10 = 80 \, \text{s}$$

Answer:

- S = 1430 m

- t = 80 s = 1 m 20 s

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