The preliminary design for a rapid-transit system calls for the train velocity to vary with time

step 1 In this problem we have a given figure here. So as shown, and the given distance from point A to

The preliminary design for a rapid-transit system calls for...

Key Concepts

Kinematics

Kinematics is the study of motion without regard to the forces that cause it, focusing on parameters such as displacement, velocity, and acceleration. In problems of motion planning, these relationships are used to design systems that meet spatial and temporal constraints, ensuring efficient and safe transit between points.

Cubic Polynomial Trajectories

Cubic polynomial trajectories are a method for modeling motion using third?degree polynomials. With the general form of a + bt + ct² + dt³, these trajectories are popular in engineering applications because they can be designed to satisfy specific conditions, such as ensuring that the velocity and acceleration profiles are continuous and smooth.

Boundary Conditions and Smooth Transitions

In motion design, imposing boundary conditions (like requiring zero slopes at the endpoints of a cubic curve) ensures smooth transitions between different motion phases. This minimizes abrupt changes in acceleration, often referred to as jerk, which is critical for maintaining passenger comfort and mechanical integrity in transit systems.

Calculus in Motion Analysis

Calculus plays a central role in motion analysis, where differentiation is used to obtain velocity and acceleration from displacement functions, and integration is applied to determine displacement from velocity functions. This mathematical framework enables the precise design and analysis of motion profiles that must adhere to specific constraints over time and distance.

Transcript

00:07 In this problem we have a given figure here.

00:11 So as shown, and the given distance from point a to b, this distance is 3 ,200 meter between stations a and b.

00:30 And the slope of the cubic transition curve, this from point o, highway from zero to this point and from this point to this point.

00:45 So the function has an equation a plus b, bt plus ct squared, plus dt cube.

00:59 So we have a given condition here from the problem.

01:05 The curve, i mean the slope of the two transition curves at zero and at end points, are zero.

01:16 So what does it mean? the slope of this curve, this one, and this another curve, at this 2 .0 and this at 15 seconds, so they are zero.

01:34 So that is the given condition.

01:39 So let's write it here.

01:45 Given conditions.

01:50 Given conditions from the problem.

01:53 So when time is equal to zero, so i'll try to make, okay, when time is equal to zero, then acceleration is equal to zero because the slope of the velocity versus time curve is the acceleration.

02:23 So when it says there the problem, in the problem that the slopes are zeros, it means that the acceleration is zero.

02:31 So at t is equal to 15 seconds, so acceleration also is zero because the slope is still zero.

02:42 The next condition is at time is equal to zero, velocity is equal to zero also.

02:52 So at this point, at this point, so we have a zero velocity, but after 15 seconds, at t is equals to 15 seconds, velocity now is equal to 36 .11 meter per second.

03:16 So this 36 .11 meter per second is just a converted value of this 130 kilometer per hour.

03:27 So we have 36 .11 meter per second.

03:29 So because from the given problem, velocity is 10030 kilometers per hour.

03:29 So we have 36 .1 meter per second.

03:34 So because from the given problem, velocity is 30 kilometer per hour.

03:44 So given that we have in one hour we have 3 ,600 seconds and also within one kilometer we have 1 ,000 meter.

04:00 So the resulting value is 36 .11 meter per second.

04:10 So we have this velocity.

04:28 The next thing we are going to do is to determine the total runtime between the station and the maximum acceleration.

04:41 So that is what is asking the problem.

04:44 The total runtime, so the total runtime of the transit from point a to point b.

04:50 So we have from this, say, let's say, this one is point c.

05:01 We'll assume that this one is point c.

05:04 And this one is point d okay so from a to c the total runtime is 15 seconds and from d to b the total runtime is already given 15 seconds so runtime from c to d is unknown so we have a given equation for velocity so therefore we can determine the distance covered first before we can determine the time of travel.

05:34 So we have a given velocity equation.

05:48 So from this formula, in order to determine the time, velocity is equals to distance traveled over time.

05:58 So in order to determine the time, so because we are asked to determine the total runtime, so distance over the velocity.

06:09 We are given with the function of of this velocity and then in the form of a plus b t squared bt plus c t squared plus d t cube so this is this was already given and the s is unknown so before we determine t we have to determine first the variable s or the total distance or the distance traveled distance traveled from point a to point c distance travel from point c to d and also from b to b.

06:49 Okay so let's proceed from the given equation here's the solution so from the given equation we have the equation of velocity which is equivalent to a plus b t plus c t squared plus d t cube so get get a derivative of that velocity with respect to time in order to determine the equation for acceleration.

07:30 So we have dv over d t is equals to b.

07:35 So we'll just remove that a because that has no variable t.

07:42 So plus 2c t plus 3d t squared.

07:51 So therefore equation for axel acceleration is b plus 2c t plus 3d t square.

08:01 This one is the equation for velocity and this one is the acceleration equation.

08:33 So we have four conditions to use.

08:36 We have this four conditions we can use in determining the variables a, b, c, and d.

08:45 So let's start with acceleration.

08:48 So with acceleration using this acceleration equation.

08:54 So when t is equal to zero, acceleration is zero.

09:03 So plug in the values here.

09:08 C, acceleration is zero.

09:11 B plus 2c, t is zero, plus 3d, t is zero.

09:19 Again squared so cancel this out therefore b is equal to zero so we already know the value of variable b so how about the other variable okay for next condition when t is equal to 15 seconds then acceleration is equal to zero so acceleration is equal to to zero.

09:55 From that formula again, 0 is equal to b is 0 plus 2c, t is 15 plus 3d, t is 15 raised to 2.

10:14 So therefore, 0 is equal to 30c plus 675d.

10:29 So we already have an equation.

10:34 So let's denote this as equation one.

10:41 Equation one.

10:43 So for acceleration, i for velocity equation, we can use also the two conditions here.

10:53 So at t is zero, velocity is zero, at t is 15, velocity is 36 .11 meter per second.

11:02 So let's proceed.

11:06 Use.

11:07 Velocity so when t is equal to zero velocity is equal to zero so we are going to use the equation for velocity of course so velocity is a plus b t plus c t squared plus d t cube so velocity is zero a this one zero t is zero plus a c t is zero again plus d t is zero again raised to three so we can cancel this out so giving us the value of a which is equivalent to zero so the next condition last one when okay when t is equals to 15 seconds velocity now becomes 30 point 11 meter per second.

12:14 So as we can see here in the conditions given from the problem.

12:21 So let's proceed.

12:28 V is equal to a plus bt plus c t squared plus d t cube.

12:36 Okay, velocity is 36 .11 meter per second.

12:45 Goes to a.

12:46 So the value of a here is zero.

12:48 So therefore that is equivalent to zero.

12:50 The value of b is 0 also, 0 multiplied by 15, plus c is unknown yet, and t is 15 squared plus t multiplied by t that is 15, so raise to 3.

13:12 So let's cancel this out because they're all zero.

13:16 So what's left is 36 .11 meter per second is equal to 15 squared is equivalent to 225.

13:31 225c plus 15 raise to 3 is 3375d.

13:39 So we already have the third equation, i mean the second equation.

13:44 So equation number 2.

13:50 So from this two equation, equation 1 and equation 2.

13:56 So from equation 1 and equation 2, equation 1 and equation 2, we have two equations and 2 unknowns also.

14:11 We have c and d.

14:13 So from this, we can derive the value of c and d.

14:22 So to equation 2, known get the value of c and d we have 0 .4814 and the value of d is negative 0 .0213.

14:47 So now we can put the variables, the value of each variables in the equation, equation for velocity and equation for acceleration.

14:57 So for equation of acceleration, so i'll just put it here, the other side.

15:10 For equation of acceleration, we have a is equals to b plus 2ct plus 3 t squared.

15:31 So value of b is 0 and c has already the value.

15:38 So we can now substitute the value for acceleration.

15:49 We have a is equal to 2 multiplied by the value of c is 0 .4814t plus 3...

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