Question

a train at station p accelerates uniformly from rest until it attains a speed of 100km/h. It then continues at that speed for some time and decelerates uniformly until it comes to a stop area station, Q 60km from P. The total time taken for the journey is one hour. If the rate of deceleration is twice that of the acceleration, calculate the (I) Time taken during which the constant speed is maintained. (II) Acceleration of the train.

Answer 1

Answer on Question #82105 Physics / Other

A train at station p accelerates uniformly from rest until it attains a speed of $100\,\mathrm{km/h}$. It then continues at that speed for some time and decelerates uniformly until it comes to a stop area station, Q 60 km from P. The total time taken for the journey is one hour. If the rate of deceleration is twice that of the acceleration, calculate the (I) Time taken during which the constant speed is maintained. (II) Acceleration of the train.

Solution:

Let $t_1$ is the time of acceleration, $t_2$ is the time of moving with constant speed, $t_3$ is the time of deceleration.

$$t_1 + t_2 + t_3 = 1\,\mathrm{h}$$

Since the rate of deceleration is twice that of the acceleration, we get $t_3 = \frac{1}{2} t_1$. So

$$1.5t_1 + t_2 = 1\,\mathrm{h}$$

The total distance traveled by the train

$$s_1 + s_2 + s_3 = 60\,\mathrm{km}$$

where

$$s_1 = \frac{0 + v}{2} t_1 = 50 t_1, \quad s_2 = v t_2 = 100 t_2, \quad s_3 = \frac{v + 0}{2} t_3 = 50 t_3 = 25 t_1$$

Thus

$$50 t_1 + 100 t_2 + 25 t_1 = 60\,\mathrm{km}$$ $$75 t_1 + 100 t_2 = 60\,\mathrm{km}$$

Equations (1) and (2) have solution

$$t_1 = \frac{8}{15}\,\mathrm{h}, \quad t_2 = \frac{1}{5}\,\mathrm{h} = 12\,\mathrm{m}$$

Acceleration

$$a = \frac{\Delta v}{t_1} = \frac{100\,\mathrm{km/h}}{\frac{8}{15}\,\mathrm{h}} = 187.5\,\mathrm{km/h^2}$$

Answer: $\frac{1}{5}\,\mathrm{h} = 12\,\mathrm{m}, 187.5\,\mathrm{km/h^2}$

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