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| Subject:
Motion problem
Category: Science > Physics Asked by: meganerd-ga List Price: $15.00 |
Posted:
20 Sep 2004 18:36 PDT
Expires: 20 Oct 2004 18:36 PDT Question ID: 404018 |
An object at rest at point d0 is accelerated at rate a until the
velocity reaches velocity v1 at point d1. Velocity remains constant
until d2.
At point d2 the object is decelerated at rate -a to velocity v2 at
point d3. Velocity remains constant to d4
At position d4, the object is accelerated at acceleration a to d5,
then decelerated at rate -a so as to come to rest at desired position
d6 in the shortest possible time.
What is the time required to reach point d6? (as a formula)
To rephrase:
Acceleration and deceleration are at the same rate, but opposite direction.
Accelerate to velococity v1, coast to d2, decelerate to V2, coast to
d4, then come to rest at d6 as quickly as possible without exceeding
the accel/decel rate.
v1 /\
________ / \
/ \ / \
/ \____/ \
/ \
/ v2 \
/ \
d0 d1 d2d3 d4 d5 d6
Time? |
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| Subject:
Re: Motion problem
Answered By: livioflores-ga on 21 Sep 2004 08:39 PDT Rated:  |
Hi meganerd!! For references (formulas, etc.) visit the following pages: "Derivation of Kinematic Equations of Motion": http://theory.uwinnipeg.ca/physics/onedim/node7.html#SECTION00251000000000000000 "Motion in one dimension": http://physics.bu.edu/~duffy/py105/Motion1D.html -------------------- The graph: Speed v1 /\ ________ / \ / \ / \ / \____/ \ / \ / v2 \ / \ d0 d1 d2d3 d4 d5 d6 Time is a velocity-time graph. Call: D1 = d1-d0 (distance between points d1 and d0) D2 = d2-d1 (distance between points d2 and d1) D3 = d3-d2 (distance between points d3 and d2) D4 = d4-d3 (distance between points d4 and d3) D5 = d5-d4 (distance between points d5 and d4) D6 = d6-d5 (distance between points d6 and d5) The distances Di are needed inputs like V1 and V2. Time for D1: (t1) V1 = a*t1 then: t1 = V1/a ----------------- Time for D2: (t2) D2 = V1*t2 Then t2 = D2/V1 ------------------ Time for D3: (t3) V2-V1 = -a*t3 then t3 = (V1-V2)/a -------------------- Time for D4: (t4) D4 = V2*t4 Then t4 = D4/V2 -------------------- Time for D5: (t5) Call V5 the final speed at d5, we have: V5^2 = V2^2 + 2*a*D5 then V5 = sqrt(V2^2 + 2*a*D5) V5-V2 = a*t5 then t5 = (V5-V2)/a = (sqrt(V2^2 + 2*a*D5) - V2) / a -------------------- Time for D6: (t6) 0 - V5 = -a*t6 then t6 = V5/a = sqrt(V2^2 + 2*a*D5) / a --------------------- The total time T is: T = t1 + t2 + t3 + t4 + t5 + t6 = = V1/a + D2/V1 + (V1-V2)/a + D4/V2 + t5 + t6 = t5 + t6 = (V5 - V2)/a + V5/a = (2*V5 - V2)/a = = (2*sqrt(V2^2 + 2*a*D5) - V2) / a then T = V1/a + D2/V1 + (V1-V2)/a + D4/V2 + (2*sqrt(V2^2 + 2*a*D5) - V2)/a = = [V1 + V1 - V2 + 2*sqrt(V2^2 + 2*a*D5) - V2]/a + D2/V1 + D4/V2 = = [2*V1 - 2*V2 + 2*sqrt(V2^2 + 2*a*D5)]/a + D2/V1 + D4/V2 = = 2*[V1 - V2 + sqrt(V2^2 + 2*a*D5)]/a + D2/V1 + D4/V2 ----------------------------------------- I hope that this helps you. If you find that something is unclear, need further assistance or if I misunderstood the problem don't hesitate to request for an answer clarification; I will gladly respond your requests. Best regards. livioflores-ga | |
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meganerd-ga
rated this answer:
and gave an additional tip of:
$10.00
My statement of the problem was not as clear as it should have been. I'm still hoping to see the final equation. |
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| Subject:
Re: Motion problem
From: hfshaw-ga on 21 Sep 2004 21:40 PDT |
As I understand it, you want to answer the question: "Given a starting
velocity (v2), how long (t_a) should I accelerate at a constant
acceleration (a), and how long (t_d) should I then decelerate at a
constant negative acceleration (-a), if I want to travel a specified
distance (D = d6 - d4) and have zero velocity when I have finished
traveling that distance?"
You have phrased the problem in terms of the distance, d5, but if we
know the t_a and the starting velocity, we can always calculate the
distance travelled.
In addition, you have phrased the problem as though it is an
optimization problem, in which you are able to choose how long to
accelerate in order to minimize the total time. In fact, given an
initial velocity, a distance that must be traveled, and an
acceleration, there is only one (or zero) possible solutions to this
problem. There are no solutions for cases in which the initial
velocity, v2 > (2*D*a)^1/2. When v2<=(2*D*a)^1/2, there is exactly
one solution.
Let:
t_a be the time during which one accelerates
t_d be the time during which one decelerates
v2 be the initial velocity (at d4 in your diagram)
v5 be the velocity after accelerating for a time t_a (the velocity at
the point d5 in your diagram)
v6= 0 be the velocity at d6
D = d6 - d4 be the distance that must be covered.
We have that:
v5 = v2 + a*t_a
v6 = 0 = v5 - a*t_d
Eliminating v5 between these two equations yields
0 = v2+a*t_a - a*t_d
Rearranging to solve for t_d yields:
t_d = (v2 + a*t_a)/a
We also know that:
D = v2*t_a + 1/2*a*(t_a)^2 + v5*t_d - 1/2*a*(t_d)^2
where the first two terms on the r.h.s. give the distance travelled
during the acceleration phase, and the last two terms give the
distance travelled during the deceleration phase.
Substituting the above expressions for v5 and t_d into the distance
equation yields (after a little algebra):
D = 2*v2*t_a + ((v2)^2)/(2*a) + a*(t_a)^2
Using the quadratic formula to solve for t_a yields:
t_a = {-v2 +/- 1/2*(2*(v2)^2 + 4*D*a)^1/2}/a
because t_a must be positive, we can immediately eliminate the root
involving the negative sign.
So, in order to meet the constraints you have specified, one should accelerate for
t_a = {-v2 + 1/2*[2*(v2)^2 + 4*D*a]^1/2}/a
Note that only solutions for which t_a> 0 are physically reasonable
(i.e., we can't accelerate for a negative amount of time once we're at
point d4). That means that:
0<= {-v2 + 1/2*[2*(v2)^2 + 4*D*a]^1/2}/a
Solving this for v2 yields the constraint v2 <= (2*D*a)^1/2 that I
mentioned at the beginning of this comment.
For your case, in which (s stands for sample):
a=0.175 lines/s^2
v2=40 lines/s
d4=170,000 lines
d6=240,000 lines
We have that:
D = 70,000 lines
t_a = 424.209 s
d5 = 32714.3 lines
t_d = 652.781 s |
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