Determining Travel Time Across Multiple Sections

Question

I need to know how to calculate the total time traveled by a vehicle traveling in a straight line split in 3 sections. First section is 6.1 ft, second 6.2 ft and third 24.1 ft. The vehicle has a max speed of 660 feet/min and acceleration rate of 5133.33 feet per squared minute and deceleration rate of -13933.3 feet per squared minute. The initial velocity = 0 and the end velocity = 0.  The equipment never reaches max speed so it must accelerate until it needs to decelerate (slow down) as to not overshoot the destination. I need to know how to calculate the time for each section.  Based on what I think I know the equipment will accelerate into the 3rd section before it needs to slow down.  I can calculate using the entire distance and I get a result of .1393 minutes traveled but I need to be able to break this down by section.

 

 

 

Answer 1

For this problem, we know that the vehicle starts and ends at rest, covers 36.4 ft total, and speeds up for part of that time, and slows down for the rest.  Using the subscript 1 to indicate the speeding up segment, and 2 to indicate slowing down, let's summarize what we know:

 

a₁ = 5133.33 ft/min²

a₂ = -13933.3 ft/min²

v_o1 (the initial velocity for speeding up) = v₂ (the final velocity for slowing down) = 0

v₁ (the final velocity for speeding up) = v_o2 (the initial velocity for slowing down)

d₁ (the speeding-up distance) + d₂ = 36.4 ft

 

To frame the whole motion, we need to figure out how far it goes while speeding up (solve for d₁).  We can do that by combining all of the things we know into a couple standard kinematic equations:

 

v₁² = v_o1²+2a₁d₁

v₂² = v_o2²+2a₂d₂

 

Since v₁ = v_o2, we can combine these into one equation, and insert all the quantities we know:

 

v₂² = v₁²+2a₂d₂ --> v₂²=v_o1²+2a₁d₁+2a₂d₂ --> 0² = 0²+2a₁d₁+2a₂d₂ --> a₁d₁= -a₂d₂ --> a₁d₁ = -a₂(36.4-d₁)

 

which, putting in values we know, becomes:

 

5133.33(d₁) = -(-13933.3)(36.4-d₁) = 13933.3(36.4-d₁)

 

With some algebra, we can solve this equation for d₁, the speeding-up distance.  I obtained d₁ = 26.6 m, making the slowing down distance d₂ = 9.8 m.

 

This means that, for the distance segments listed in the problem, the vehicle is speeding up through the entire first two segments and for over half the final segment.  I will label those segments a, b, and c, in the order presented in the question.

 

Now we can figure out the time for each segment using other kinematic equations.  Sanhita already did this for the first two in the same way I will, but I will include them here for completeness.

 

To solve for t_a, we can use:

 

d_a = v_oat_a + (1/2)a₁t_a² --> 6.1 = 0 + (1/2)(5133.33)t_a² 

 

which gives t_a = 0.049 min

 

Since we always need to know the initial velocity for any motion to accurately determine other quantities later, we need to solve for v_ob before moving on.  Since v_a = v_ob,  We can use:

 

v_a = v_o+a₁t_a = v_ob

 

This gives v_ob = 0+(5133.33)(0.049) = 251.5 ft/min

 

Then we can repeat for Segment b (the 6.2-ft segment), where the vehicle is still only speeding up.

 

d_b = v_obt_b+(1/2)a₁t_b²  --. 6.2 = (251.5)t_b+(1/2)(5133.33)t_b²

 

This can be rearranged into a standard quadratic equation in t_b.  I used the quadratic formula to solve it, and obtain:

 

t_b = 0.02 min (the positive-time solution, as Sanhita pointed out, since our physical time here cannot be negative)

 

Then, solve for the initial velocity of Segment c = final velocity of Segment b.

 

v_b = v_ob+a₁t_b = 251.5 + (5133.33)(0.02) = 354.2 ft/min

 

Now, in Segment c, the vehicle is slowing down for the last 9.8 ft, which means it is spending the first 24.1 - 9.8 = 14.3 ft speeding up.  We can thus determine the time for each of the speeding up and slowing down parts in this segment, and then add those times together to get the total time spent in segment C (using the final speeding up velocity as the initial slowing down velocity). 

 

For Segment c speeding up:

 

d_cu = v_ocut_cu+(1/2)a₁t_cu² --> 14.3 = 354.2t_cu+(1/2)(5133.33)t_cu²

 

Again, this becomes a quadratic to solve for t_cu.  When I do that, I get t_cu = 0.033 min as the positive-time solution.

 

Solving for the final speeding up velocity, which is the initial slowing down velocity (v_cdo):

 

v_cu = v_cuo+a₁t_cu = v_cdo = 354.2+(5133.33)(0.033) = 523.6 ft/min  (not exceeding 660 ft/min, so an okay speed)

 

Then, finally, for the slowing down segment, since we know the final speed v_cd = 0, we can use a slightly different kinematic expression to solve for time:

 

d_cd = (1/2)(v_cdo+v_cd)t_cd --> 9.8 = (1/2)(523.6+0)t_cd

 

This gives t_cd = 0.037 min.

 

The total time spent in the last 24.1 ft is thus t_cu+t_cd = 0.033 + 0.037 = 0.07 min.

 

I did not recheck all these calculations, so no guarantees about errors.  But my results match Sanhita's, as far as they are the same (until Segment c).  But the technique should be sound.  Just let me know if you have any questions about details or algebra.  Hope this helps!  And thanks for clearing up the statements of the problem.

 

Answer 2

Let's assume that the vehicle traveling in a straight line split in 3 sections traveled as follows:

1. First section of 6.1 ft distance (S₁) in t₁ minutes from start
2. Second section of 6.2 ft distance (S₂) in t₂ minutes from end of S₁ and 
   
3. Third section of 24.1 ft distance (S₃) in t₃ minutes from end of S₂ 
4. Time required to attain maximum velocity in t minutes

And also assume that it starts from rest and ends at rest, end points of S₁ and S₂ are respectively starting points of S₂ and S₃ so that the sections are continuous and and there exists no gap  

 

Given that, the maximum speed that can be attained by the vehicle v_max is=  660 feet/min, 

acceleration, a = 5133.33 feet per squared minute

deceleration, a' = -13933.3 feet per squared minute

and initial velocity, v₀ = 0

We have, t = [v_max -v₀]/a = [660-0]/5133.33 ≈ 0.13 minute

Also we have, S₁= v₀t₁ + 0.5at₁² 

Substituting the values, 6.1=0+0.5*5133.33t₁² 

=> 2566.67t₁² =6.1 

=>t₁² = 6.1/2566.67

=>t₁² ≈ 0.0024

=>t₁≈0.05 minute which is less than t and the vehicle may accelerate over following t' minutes = (t - t₁) minutes = [0.13-0.05] minutes =0.08 minutes

 

At end point of S₁ velocity of the vehicle, v' = v₀ + at₁ = 0+ 5133.33* 0.05 = 256.67 feet per minute

Thus, for t' =0.08 minutes the vehicle accelerates and then decelerates.

And the vehicle in  t' =0.08 minutes covers a distance, S'= v't' + 0.5at'² = 256.67*0.08+0.5*5133.33*(0.08)² ≈ 20.53+ 1.64 ≈ 22.18 feet which is greater than S₂ 

Therefore t₂ < t'

Hence, S₂= v't₂ + 0.5at₂² 

Substituting the values, 6.2 = 256.67 t₂ + 2566.67t₂² 

=> 2566.67t₂² + 256.67 t₂ -6.2 =0 

=> t₂ = {- 256.67 ±√[(256.67)² - 4* 2566.67*(-6.2)]}/ (2*2566.67)={- 256.67 ±√[65879.49 + 63653.42]}/5133.33 ={- 256.67 ±√[129532.9089]}/5133.33={- 256.67 ±359.91}/5133.33 =103.24/5133.33 or (-616.58)/5133.33

Since time cannot be negative, t₂ ≈ 0.02 minutes which is less than t' and the vehicle may accelerate over following t" minutes = (t' - t₂) minutes = [0.08-0.02] minutes =0.06 minutes

 

At end point of S₂ velocity of the vehicle, v" = v' + at₂ = 256.67+ 5133.33* 0.02 ≈ 256.67 + 102.67 ≈ 359.34 feet per minute
Thus, for t" =0.06 minutes the vehicle accelerates and then decelerates.
And the vehicle in t" =0.06 minutes covers a distance, S"= v"t" + 0.5at"² = 359.34*0.06+0.5*5133.33*(0.06)² ≈ 21.56+ 9.24 ≈ 30.80 feet which is greater than S₃

 

Therefore t₃ < t"
Hence, S₃= v"t₃ + 0.5at₃² 

Substituting the values, 24.1 = 359.34t₃ + 0.5*5133.33 (t₃)² 

=> 2566.67t₃² + 359.34 t₃ -24.1 =0

=> t₃ = {-359.34±√[(359.34)² - 4*2566.67(-24.1)]}/(2*2566.67) ≈ {-359.34±√[129125.24 +247426.99 ]}/5133.33 ≈ {-359.34±√[376552.23]}/5133.33 ≈ {-359.34±613.64}/5133.33 ≈ 254.3/5133.33 or (-972.3)/5133.33

Since time cannot be negative, t₃ ≈ 0.05 minutes

 

Therefore, starting from rest, the travel time of the vehicle is almost 0.05 minutes for the first section, nearly 0.07 minutes for the second section and approximately 0.12 seconds for the third section.

Answer 3

To avoid tutors from clicking on a problem and finding that another tutor is awaiting more information, perhaps WyzAnt could show "# of comments" at the top level where all the questions are listed.

 

I am making a comment here instead of an "answer" so tutors will not waste time.