Question

The maximum possible acceleration of a train starting from the rest and moving on straight track is $10m/s^2$ and maximum possible retardation is $5m/s^2$. The maximum speed that train can achieve is $70m/s$. Minimum time in which the train can complete a journey of $1000m$ ending at rest is $\frac{347}{2\alpha }sec$. Where $\alpha$ is an integer. Find $\alpha$.

Answer 1

Distance cover when speed becomes maximum $=70m//s$

$v^2-u^2=2a{x}_1$

$70\ast 70=2\ast 10x_1$

$x_1=\frac{70\ast 70}{5\ast 2}=245m$

time taken $t_1=\frac{70}{10}=7sec$

Distance cover retadation from speed 70 ms to 0

$0-{70}^2=-2\ast 5x_3$

$x_3=\frac{70\ast 70}{5\ast 2}=490$

Time taken $t_3=\frac{70}{5}=14sec$

$t_2=\frac{265}{70}=\frac{53}{14}$

minimum time $t=t_1+t_2+t_3=7+14+\frac{53}{14}$

$=\frac{347}{14}=\frac{347}{2\alpha }$

$\alpha =7$