Question

The greatest possible acceleration or deceleration that a train may have is $a$ , and its maximum speed is $v$ and total distance is $s$. What is the minimum time in which the train can get from one station to another?

• A. $t=\frac{s}{v}+\frac{v}{a}$
• B. $t=\frac{s}{v}$
• C. $t=\frac{s}{v}-\frac{v}{a}$
• D. $t=\frac{s}{v}+2\frac{v}{a}$
  

Answer 1

The correct option is A

$t=\frac{s}{v}+\frac{v}{a}$

Maximum acceleration and deceleration is same. So time for reaching the maximum speed is same as time for reaching the zero speed from maximum speed.

$t_1=\frac{v}{a}=t_3$

$s=ut+\frac{1}{2}a{t}^2$

$s_1=s_3=\frac{1}{2}\times a\times \frac{v^2}{a^2}$

$s_1=s_3=\frac{v^2}{2a}$

$s_2=v\times t_2$

$t_2=\frac{s_2}{v}=\frac{s-2\times \frac{v^2}{2a}}{v}$

$t_2=\frac{s}{v}-\frac{v}{a}$

Now, total time $t=t_1+t_2+t_3$

$=\frac{v}{a}+\frac{s}{v}-\left(\frac{v}{a}\right)+\left(\frac{v}{a}\right)$

$t=\frac{s}{v}+\left(\frac{v}{a}\right)$