# Kinematics, Distance acceleration and deceleration.

## Homework Statement

There are two train stations, 2.5km (2500m) apart, therefore an ordinary suburban train cannot reach its max v between the two station. They try to minimise time taken by accelerating at a constant rate of 0.1ms-2 for time t1 and decelerate at 0.5ms-2 for time t2.

Find the maximum speed of the train, and the minimum time taken to travel this distance.[/B]

a = v/t
v = d/t

## The Attempt at a Solution

I'm not so much stuck on the side of figuring out max velocity and time, I for some reason can not for the life of me figure out how I'm supposed to determine the DISTANCE it takes to reach the max speed.

I know the ratio of acceleration is 1:5, but can't figure out how this then relates into the distances required for acceleration and breaking. I'm not after a full answer because this is an assignment question, but more a poke in the right direction.

Thank you

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Welcome to PF. Love your name. Could you post your work? I can't do much to help you if I cannot see your work. Additionally, I would recommend you take a look at the introductory physics formulae thread. You are missing a crucial bit of information.

Delta2
Homework Helper
Gold Member
Basically you seem to neglect the equation that relates the distance traveled to the time it takes during acceleration or deceleration. It is $s_1=0.5a_1t_1^2$ and $s_2=v_it_2-0.5a_2t_2^2$ where v_i the velocity at the time t1 where acceleration stops and deceleration starts. According to the constrains given by the problem it should be $s_1+s_2=2.5km$ and $v_i-a_2t_2=0$

Hello Aleph and Delta, thanks for your replies, sorry I didn't get back sooner, I was grinding away at this question and I think I have it but would like to let you two say what you think before I go ahead and submit it.

Known:

a1 = 0.100ms-2
a2 = 0.500ms-2
dtotal = 2.5km = 2500m

vi = vf = 0ms-1

Equations:

v1 = a1t1 = -a2t2
d = ½ at2

Solution

d = ½ a1t12 + v1t2 + ½ a2t22

2500m = ½ a1t12 + a1t1 (a1t1 / a2)2
= ½ a1 (1 – a1 / a2) t12

(1 – a1 / a2)t12 = 2500/(½a1)
(1 – a1 / a2)t12 = 50000
t12 = 50000 / (1 – a1/a2)
= 50000 / (1 – 0.100ms-2 / 0.500ms-2)
= 62500
t1 = √62500
= 250s

t2 = a1t1 / a2
= 0.100ms-2 • 250s / 0.500ms-2
= 50s

ttotal = t1 + t2
= 300s

PeroK
Homework Helper
Gold Member
If you calculate the distance travelled during ##t_1 = 250s## you will see it's more than 2500m.

Note also that:

As the deceleration is 5 times the acceleration, you can see that ##t_1 = 5t_2##.

Can you use that to find the relationship between the distances ##d_1## and ##d_2##?

Last edited:
Quite right Perok!

Redoing my work after your find I have reached the following conclusion that checks out when plugged into vf = vi + at

t1 = 200s
t2 = 40s

40s also happens to be one fifth of 200s, making it such a lovely fit.

Thank you all so much for your help.

PeroK
Homework Helper
Gold Member
Quite right Perok!

Redoing my work after your find I have reached the following conclusion that checks out when plugged into vf = vi + at

t1 = 200s
t2 = 40s

40s also happens to be one fifth of 200s, making it such a lovely fit.

Thank you all so much for your help.
That's the solution for 2.4km.