Question

a train starts its journey with constanttacceleration alpha ,attains a velocity v and then moves with v for some distance and then deaccelerates at the rate of beta to come to rest .if the total length of the path covered is L, find out the total time interval of motion

Answer 1

a train starts its journey with constant acceleration alpha, attains a velocity $v$ and then moves with $v$ for some distance and then deaccelerates at the rate of beta to come to rest. If the total length of the path covered is $L$, find out the total time interval of motion

1. Acceleration:

$$v = \alpha t _ {a}$$

$t_a$ – time of acceleration

$$t _ {a} = \frac {v}{\alpha}$$

Distance travelled:

$$l _ {a} = \frac {\alpha t _ {a} ^ {2}}{2} = \frac {v ^ {2}}{2 \alpha}$$

2. Uniform motion:

Distance travelled:

$$l _ {u} = v * t _ {u}$$

$t_u$ – time of uniform motion

3. Deceleration

$$v = \beta t _ {d}$$

$t_d$ – time of deceleration

$$t _ {d} = \frac {v}{\beta}$$

Distance travelled:

$$l _ {d} = \frac {\beta t _ {d} ^ {2}}{2} = \frac {v ^ {2}}{2 \beta}$$

Total time equals:

$$t = t _ {a} + t _ {u} + t _ {d} = \frac {v}{\beta} + \frac {l _ {u}}{v} + \frac {v}{\alpha}$$

Total distance:

$$l = l _ {a} + l _ {u} + l _ {d}$$

Therefore: $l_{u} = l - l_{a} - l_{d} = l - \frac{v^{2}}{2\beta} - \frac{v^{2}}{2\alpha}$

$$t = \frac {v}{\beta} + \frac {l _ {u}}{v} + \frac {v}{\alpha} = \frac {v}{\beta} + \frac {l - \frac {v ^ {2}}{2 \beta} - \frac {v ^ {2}}{2 \alpha}}{v} + \frac {v}{\alpha} = \frac {v}{2 \beta} + \frac {l}{v} + \frac {v}{2 \alpha}$$

Answer: $t = \frac{v}{2\beta} + \frac{l}{v} + \frac{v}{2\alpha}$