max. speed and acceleration

Answer
It is an incomplete information ... Whithout some info, it seems not possible to answer the question.

BUT we can try answering your question ...

Point 1. What length of the train? Normally 150 meters or more in metropolitan cities

Point 2. What part of the train is 30 m ahead of you when you reach the station? Is it the last Berth (tail) of the train or is it the engine (head) ?

But as per you question you have to do a last moment jump into the train .... So it is most probably tail of the train ahead of you.

Becoz if the engine of the train was ahead then you could have very well jumped into some bogey of the train at that moment considering 120 meters still behind you ...

Point 3: What is the acceleartion of the train which has started to speed up? Slowest would be 1 m/s^2



CASE 1:

But we can make asssumptions and try to answer this question

Point 1: Assume the tail of the train is 30 m ahead of you
Point 2: Assume Length of the train is 150 meters
Point 3: Acceleration of the train is 1 m/s^2


Lets solve

Train has just started with 0 m/s

Assume your velocity is constant (there is not acceleration from your side)

You have 50 m to cover before reaching barrier or jump into the train and your beginning velocity is 0.8 m/s.

So maximum time taken to cover distance of 50m by you before you falling into the barrier = 50/0.8 = 62.5 seconds

In those 65.5 seconds the train can move ahead by distnace s

s = ut + (1/2)at^2

s = 0*62.5 + (1/2)*1*(62.5)^2

s = 1953.125 meters

Let say s = 1953 + 30 = 1983 ... as it's tail was 30 m ahead of you when both of you started. So in time 62.5 seconds the distance is maintained as you have NOT accelerated yet.

But also in 62.5 seconds you had already covered 50 m distance so D(s) = Distance beween you and train tail = 1983 - 50 = 1933 when you were at barrier

So when you reach the barrier the TAIL of the train was already ahead of
you by 1933 meters distance

I mean if you are a spiderman or SUPER olympic long jumper, you can try jumping that 1933 meetrs onto the back step of the train

CASE 2:

To Let make you jump into the train for which we need to optimise the assumptions from CASE 1.....

Obvious step would be to make you SPEED up so Consider you starts accelerating at A m/s^2 from initial velocity 0.8 m/s


So time taken by you to cover 50 m would be

s = ut + (1/2)At^2

50 = 0.8t + (1/2)At^2

100 = 1.6*t+ At^2

At^2+1.6t-100=0

As per quadratic formula ...

t = (-1.6 +-sqrt(1.6^2-4*100*A))/2A

t = (-1.6 +-sqrt(2.56 - 400A))/2A

For the time t to be real (non imaginary)

2.56 >= 400A

2.56/400 >= A

0.0064 >= A

which shows you have to have very low acceeleration near to Zero


So going back to CASE 1 one assumption (your acceleration as Zero) is correct

Other assumptions canniot be optimised furthur

Final Conclusion:

Will you be able to leap onto the back step of the train before you crash into the barrier?

NO

What Say?

Er calculation check ... maximum speed is 8.0 metres a second, not 0.8m/sec - which a ant could do!

Recalculate! You might find its 6.25 seconds and the train covers 19.5 + 30 = 49.5 metres while you've done 50 metres in that time == success, and you're travelling only 1.75 m/s sec faster than the train at this stage so a jump in the last 0.5 metres is possible.

PS

Why do you believe the train accelerates at 1 m/s/s? And maybe you're meant to solve it for a unknown but leats say constant acceleration factor X.

OOPS Thats explains how I lost marks in my physics paper back in my school days ...

Hey, with a speed of 50m/s you have 6.25 secs b4 u crash into the board, now in this interval everything depends on the accelaration of the train, If you are in India u r done but if you are in Japan then save your teeth....

Anyways the question is not complete