Kinematics, Distance acceleration and deceleration.

[ManicPIxie]

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]

Homework Equations

a = v/t
v = d/t
v_f = v_i + 2ad

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

 

[AlephNumbers]

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]

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 $v_i-a_2t_2=0$

 

[ManicPIxie]

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]

If you calculate the distance traveled 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 $d_2$?

 

[ManicPIxie]

Quite right Perok!

Redoing my work after your find I have reached the following conclusion that checks out when plugged into v_f = v_i + 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]

Quite right Perok!

Redoing my work after your find I have reached the following conclusion that checks out when plugged into v_f = v_i + 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.