Question

the greatest acceleration or deceleration that a train may have is a the minimum time in which the train can start from one station to stop at the next ( at a straight distance of s ) is??

Answer 1

This question is very hard to understand as written, can you rephrase it?

Answer 2

The question isnt understandable.You may ask them separately.

Answer 3

So, the total distance the train has to travel is 's'. And it is to start from rest and stop at the end. The maximum acceleration as well as the maximum retardation is 'a'. And we have to find the minimum time to complete this journey. The velocity-time graph for minimum time will look like this :

What u are seeing is the train picking up speed with maximum acc 'a' and reaching maximum speed 'Vo' and then retarding at maximum at a particular moment ' t' ' and reaching to a halt at 't' . The 't' thus obtained will be minimum. For finding t ,we first need to find Vo This question is symmetric in acceleration and retardation , so we can intuitively claim that it will be at/2. But the proper procedure should be to write the following equations.

Vo = at' {eqn of motion for accelerating part of motion}
and

Vo=a(t-t') {for decelerating part of the motion}

Adding, we get Vo=at/2

We know area under v-t graph gives displacement (in this case equal to s)

So \(\large{\frac{Vo \ \times t}{2}}\) =s

This gives \(\large{t=2\sqrt{s/a}}\)

This is the minimum time for the journey.