Question

A train at station P accelerates uniformly from rest until it attains a speed of 100 km/h. It then continues at the speed for some time and decelerates uniformly until it come to a stop at a station Q 60 km from P. The total time taken for the journey is on hour. If the rate of deceleration is twice that of the acceleration. Calculate the:

(i) Time taken during the constant speed is maintained

(ii) Acceleration of the train

Answer 1

Answer on Question #82082, Physics / Mechanics (fix 1)

A train at station P accelerates uniformly from rest until it attains a speed of $100\,\mathrm{km/h}$. It then continues at the speed for some time and decelerates uniformly until it comes to a stop at a station Q 60 km from P. The total time taken for the journey is on hour. If the rate of deceleration is twice that of the acceleration. Calculate the:

(i) Time taken during the constant speed is maintained

(ii) Acceleration of the train

Solution

$$a = \frac{v - v_0}{t}$$ $$a_1 = \frac{100 - 0}{t_1} \quad a_3 = \frac{0 - 100}{t_3}$$ $$a_2 = 2a_1$$ $$\frac{200}{t_1} = \frac{-100}{t_3}$$ $$t_1 = 2t_3$$

From the condition: $\left\{ \begin{array}{c} t_1 + t_2 + t_3 = 1 \\ t_1 v_{average1} + t_2 v_{average2} + t_3 v_{average3} = 60 \end{array} \right\}$

$$\left\{ \begin{array}{c} 2t_3 + t_2 + t_3 = 1 \\ 50t_1 + 100t_2 + 50t_3 = 60 \end{array} \right\}$$

Solving the system of equations we obtain:

$$t_1 = \frac{8}{15} \text{ hour}$$ $$t_2 = \frac{3}{15} \text{ hour}$$ $$t_3 = \frac{4}{15} \text{ hour}$$ $$a _ {1} = \frac {v - v _ {0}}{t _ {1}} = \frac {1 0 0 \text { kilometer per hour}}{\frac {8}{1 5} \text { hour}} = \frac {\frac {1 0 0 0 0 0 \text { meter}}{3 6 0 0 \text { sec}}}{1 9 2 0 \text { sec}} = 0. 0 1 4 \frac {\text { meter}}{\text { sec} ^ {2}}$$

Answer: $t_2 = \frac{3}{15}$ hour; $a_1 = 0.014\frac{\text{meter}}{\text{sec}^2}$

Answer provided by https://www.AssignmentExpert.com