Question

the motors of an electric train can give it an acceleration of 1 m/s2 and the brakes can give a negative acceleration of 3m/s2. find the shortest time interval in which the train can make a journey between two stations 1210m apart .

Answer 1

Question#20663

the motors of an electric train can give it an acceleration of $1 \, \text{m/s}^2$ and the brakes can give a negative acceleration of $3 \, \text{m/s}^2$. find the shortest time interval in which the train can make a journey between two stations 1210m apart.

Solution:

Let:

$$S = 1210 \, \text{m}$$ $$a_1 = 1 \, \text{m/s}^2$$ $$a_2 = 3 \, \text{m/s}^2$$

t-?

$$S = S_1 + S_2$$

Were:

$S_1$ is the distance of moving with the positive acceleration

$S_2$ is the distance of moving with the negative acceleration

Such as:

$$\frac{S_1}{S_2} = \frac{a_1}{a_2} = \frac{1}{3}$$ $$S_1 = \frac{3}{4} S, \quad S_2 = \frac{1}{4} S$$ $$S_1 = \frac{1}{2} a_1 t_1^2$$ $$S_2 = -\frac{1}{2} a_2 t_2^2$$

Were:

$$t_1, t_2$$

Are the times of moving with the positive and negative accelerations respectively.

$$t = t_1, +t_2$$ $$t = \sqrt{\frac{2S_1}{a_1}} + \sqrt{\frac{2S_2}{a_2}}$$ $$\begin{array}{l} t = \sqrt{\frac{35}{2a_1}} + \sqrt{\frac{5}{2a_2}} \\ t = \sqrt{\frac{3 \cdot 1210}{2 \cdot 1}} + \sqrt{\frac{1 \cdot 1210}{2 \cdot 3}} = 56.8 \, \text{s} \end{array}$$

Answer: 56,8 s.