Need help with one dimensional kinematics problem

In summary, the author has been struggling to solve a problem for a long time and has come up with different methods, but none of which have worked. After someone provides guidance, the author solves the equations and finds that the final velocity is 0.
  • #1
azure.hubris
9
0
Hello,
I've been working on the following problem for quite some time, with no success:
A commuter train travels between two downtown stations. Because the stations are only 1.15 km apart, the train never reaches its maximum possible cruising speed. The engineer minimizes the time t between the two stations by accelerating at a rate a1 = 0.105 m/s^2 for a time t1 and then by braking with acceleration a2 = -0.585 m/s^2 for a time t2. Calculate the minimum time of travel t.​

I've approached the problem in a few different ways; I've been attempting to state the supplied information in such a way that I can treat it as a max/min optimization problem and differentiate from there. I've been setting up the information using the equation:
xfinal = xinitial + velinitial(t) + 1/2(a)t^2

for example where for the first interval t1:
xfinal = 1150m
xinitial = 0
vinitial = 0
accel = 0.105m/s^2
time = the unknown

I'm running into the conceptual problem of how to deal with the two intervals. Where the time as I've laid it out is referring to just one of the two intervals, where I need to be dealing with both, i.e. the total time. I hope someone can give me an idea of how I'm misconceiving this problem. I feel like there needs to be another equation relating the two time intervals, but I can't figure out how I need to state the variables and how to integrate both equations.
I hope I've shown enough here to incline someone to give me some guidance, because I've really run up against a brick wall on this one, and I've spent a lot of time with it.
Thank you for your time!
 
Physics news on Phys.org
  • #2
Vmax = 0.105*t1 = 0.585*t2 ------------------------(1)

1150 = 0.5*0.105*t1^2 + 0.5*0.585*t2^2 -----------(2)

Think of these two equations. can yau solve them?
 
  • #3
Thanks for your reply, but I'm not sure how to proceed, honestly. Though I do see the relation that you've made between t1 and t2 in terms of Vmax, I really don't understand why you've done this. How is knowing the max velocity pertinent?
I can work the equations, solving equation (1) for t1, subbing into equation 2, then using that result in the original equation to find the other interval, then adding the result. But that doesn't give me a correct answer, unfortunately. And ultimately, I don't understand why this method should work, which is most important to me. If someone could further clarify the methodology behind a solution, I'd really appreciate it. Thanks.
 
  • #4
can you tell us what the correct answer is ?
 
  • #5
I think it's best to set up two equations: one for the total distance traveled, and one for the final velocity of the train. You'll get:

[tex]x_{f}=\frac{1}{2}a_{1}t_{1}^{2} + v_{f,1} t_{2} - \frac{1}{2}a_{2}t_{2}^{2}[/tex]

where the subscript 'f' denotes 'final' and the subscripts '1' and '2' denote the part of the trip where the train is accelerating and the part of the trip where the train is decelerating, respectively. [tex]a_{1}[/tex] and [tex]a_{2}[/tex] are both positive in this case. Also, [tex]v_{f,1}[/tex] can be substituted with [tex]a_{1}t_{1}[/tex], so the only two unknowns left in this equation are [tex]t_{1}[/tex] and [tex]t_{2}[/tex]. The second equation, for the final velocity is:

[tex]v_{f} = a_{1}t_{1} - a_{2}t_{2}[/tex].

Of course, [tex]v_{f} = 0[/tex]. Again, in this equation, both [tex]a_{1}[/tex] and [tex]a_{2}[/tex] are taken positive. Now you have a system of two equations with two unknowns, [tex]t_{1}[/tex] and [tex]t_{2}[/tex]. This system can be solved without further problems.
 
  • #6
Actually, no. It's a problem in an online course, you enter your final numerical answer and it either accepts it as correct, or doesn't. My final answer using the method above, was 68.23 s, and I've tried interpreting it in a few different ways, arriving at different answers, but none seem to jive.
Does this method seem correct? If so, could you explain it a a little bit?
Thank you.
 
  • #7
Brinx said:
I think it's best to set up two equations: one for the total distance traveled, and one for the final velocity of the train. You'll get:

[tex]x_{f}=\frac{1}{2}a_{1}t_{1}^{2} + v_{f,1} t_{2} - \frac{1}{2}a_{2}t_{2}^{2}[/tex]

where the subscript 'f' denotes 'final' and the subscripts '1' and '2' denote the part of the trip where the train is accelerating and the part of the trip where the train is decelerating, respectively. [tex]a_{1}[/tex] and [tex]a_{2}[/tex] are both positive in this case. Also, [tex]v_{f,1}[/tex] can be substituted with [tex]a_{1}t_{1}[/tex], so the only two unknowns left in this equation are [tex]t_{1}[/tex] and [tex]t_{2}[/tex]. The second equation, for the final velocity is:

[tex]v_{f} = a_{1}t_{1} - a_{2}t_{2}[/tex].

Of course, [tex]v_{f} = 0[/tex]. Again, in this equation, both [tex]a_{1}[/tex] and [tex]a_{2}[/tex] are taken positive. Now you have a system of two equations with two unknowns, [tex]t_{1}[/tex] and [tex]t_{2}[/tex]. This system can be solved without further problems.

Thank you, thank you, thank you, Brinx! Your reply clears things up immensly, I'll work it through and see if I can get the correct answer.
 

FAQ: Need help with one dimensional kinematics problem

What is one dimensional kinematics?

One dimensional kinematics is the study of motion of objects along a straight line, without taking into consideration the causes of motion.

How do I solve a one dimensional kinematics problem?

To solve a one dimensional kinematics problem, you need to first identify the known and unknown variables, then use the equations of motion (such as velocity, acceleration, and displacement) to solve for the unknown variable.

What are the key equations used in one dimensional kinematics?

The key equations used in one dimensional kinematics are:

  • Displacement (Δx) = Average Velocity (v) x Time (t)
  • Final Velocity (vf) = Initial Velocity (vi) + Acceleration (a) x Time (t)
  • Displacement (Δx) = Initial Velocity (vi) x Time (t) + 1/2 x Acceleration (a) x Time (t)^2
  • Final Velocity (vf)^2 = Initial Velocity (vi)^2 + 2 x Acceleration (a) x Displacement (Δx)

What are some common units used in one dimensional kinematics?

Some common units used in one dimensional kinematics are:

  • Displacement: meters (m)
  • Velocity: meters per second (m/s)
  • Acceleration: meters per second squared (m/s^2)
  • Time: seconds (s)

Can you provide an example of a one dimensional kinematics problem?

Sure, here's an example: An object starts from rest and accelerates at a rate of 2 m/s^2 for 5 seconds. What is its final velocity and displacement?
Known variables: Initial velocity (vi) = 0 m/s, acceleration (a) = 2 m/s^2, time (t) = 5 s
Using the equation vf = vi + at, we can solve for the final velocity:
vf = 0 + (2 m/s^2 x 5 s) = 10 m/s
Using the equation Δx = vit + 1/2at^2, we can solve for the displacement:
Δx = (0 m/s x 5 s) + 1/2(2 m/s^2 x (5 s)^2) = 25 m
Therefore, the final velocity is 10 m/s and the displacement is 25 m.

Similar threads

Replies
4
Views
1K
Replies
20
Views
803
Replies
4
Views
1K
Replies
4
Views
836
Replies
32
Views
4K
Replies
9
Views
2K
Back
Top