Kinematics, Distance acceleration and deceleration.

  • #1
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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
vf = vi + 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
 

Answers and Replies

  • #2
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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.
 
  • #3
Delta2
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Basically you seem to neglect the equation that relates the distance traveled to the time it takes during acceleration or deceleration. It is [itex]s_1=0.5a_1t_1^2[/itex] and [itex]s_2=v_it_2-0.5a_2t_2^2[/itex] 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 [itex]s_1+s_2=2.5km[/itex] and [itex]v_i-a_2t_2=0[/itex]
 
  • #4
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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
 
  • #5
PeroK
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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:
  • #6
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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.
 
  • #7
PeroK
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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.
 

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