Question

A train starts from rest from a station with acce.02m/s2 on a straight track and than comes tto rest after attaining maximum velocity on another station due to retardation of 0.4m/s2 .if total time spent is half an hour ,than distance between two station is

Answer 1

A train starts from rest from a station with acce.02m/s2 on a straight track and then comes to rest after attaining maximum velocity on another station due to retardation of 0.4m/s2. If total time spent is half an hour, than distance between two stations is.

**Solution.**

$$a_1 = 0.2 \frac{m}{s}, \quad a_2 = 0.4 \frac{m}{s}, \quad t = 30 \text{min} = 1800 \text{s};$$ $$s - ?$$

A displacement of the train, when it went with acceleration $a_1$:

$$s_1 = \frac{v^2 - v_0^2}{2a_1}.$$

$v_0 = 0, \quad v$ - attaining maximum velocity.

A displacement of the train, when it went with retardation $a_2$:

$$s_2 = \frac{u^2 - u_0^2}{2a_2}.$$

$u = 0, \quad u_0$ - attaining maximum velocity.

A final velocity, when the train goes with acceleration $a_1$ is a initial velocity, when it goes with retardation $a_2$.

$u_0 = v$, then:

$$s_1 = \frac{v^2}{2a_1};$$ $$s_2 = \frac{v^2}{2a_2}.$$

Divide fist equation by second:

$$\frac{s_1}{s_2} = \frac{v^2 2a_2}{2a_1 v^2} = \frac{a_2}{a_1};$$ $$\frac{s_1}{s_2} = \frac{0.4}{0.2} = 2;$$ $$\frac{s_1}{s_2} = 2.$$

A velocity of the train, when it went with acceleration $a_1$:

$$v = v_0 + a_1 t_1;$$ $$v _ {0} = 0,$$ $$v = a _ {1} t _ {1}.$$

A velocity of the train, when it went with retardation $a_2$:

$$u = u _ {0} - a _ {2} t _ {2};$$ $$u = 0;$$ $$u _ {0} = a _ {2} t _ {2}$$ $$u _ {0} = v;$$ $$a _ {1} t _ {1} = a _ {2} t _ {2};$$ $$\frac {t _ {1}}{t _ {2}} = \frac {a _ {2}}{a _ {1}};$$ $$\frac {t _ {1}}{t _ {2}} = \frac {0 . 4}{0 . 2} = 2;$$ $$\frac {t _ {1}}{t _ {2}} = 2.$$

A displacement of the train (is the distance between two stations):

$$s = s _ {1} + s _ {2};$$ $$s _ {2} = \frac {s _ {1}}{2};$$ $$s = s _ {1} + \frac {s _ {1}}{2} = \frac {3}{2} s _ {1}.$$

A displacement of the train, when it went with acceleration $a_1$:

$$s _ {1} = v _ {0} + \frac {a _ {1} t _ {1} ^ {2}}{2};$$

$v_{0} = 0$ then:

$$s _ {1} = \frac {a _ {1} t _ {1} ^ {2}}{2}.$$

The total time:

$$t = t _ {1} + t _ {2};$$ $$t _ {2} = \frac {t _ {1}}{2};$$ $$t = t _ {1} + \frac {t _ {1}}{2} = \frac {3}{2} t _ {1};$$ $$\begin{array}{l} t_{1} = \frac{2}{3} t; \\ s = \frac{3}{2} s_{1} = \frac{3}{2} \frac{a_{1} t_{1}^{2}}{2} = \frac{3}{2} \frac{a_{1}}{2} \left(\frac{2}{3} t\right)^{2} = \frac{3 a_{1} 4 t^{2}}{4 \cdot 9} = \frac{a_{1} t^{2}}{3}; \\ s = \frac{a_{1} t^{2}}{3}. \\ s = \frac{0.2 (1800)^{2}}{3} = 216000 (m); \\ s = 216 km. \end{array}$$

Answer: $s = 216 \, km$.