Calculate acceleration and speed based on times

simonwait

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Jul 9, 2020
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Hi

I have been given the following information

Acceleration - 2s
Decelleration - 3s
Total Time -9.5s
Distance - 8509mm

Obviously I know that I must be at full speed for 4.5s based on the above. I need to find the acceleration, deceleration and maximum speed in mm/s (or mm/s^2). Also what about on occasions where we might never achieve full speed?

I know I have done this before but for some reason (I blame being cooped up!) I cant get my head around it.
 
Hi, do you mean it accelerated for 2 sec, then accelerated (with negative acceleration) for 3 sec and continue with constant velocity for 4.5 sec?
Are both acceleration of the same absolute value? Can you write some equation of motion that connects the total distance with the data you were given?
 
Please don't use '-' in math unless you want to use it as a negative sign, please.

Why must you be at full speed for 4.5 sec? A funny thing happened to me just yesterday. In my car I accelerated for 2sec, then immediately decelerated for 3 sec (I saw the police) and then I stayed at that speed for exactly 4.5 sec until the police were out of my site. So my slowest speed was for 4.5s.
 
Hi, sorry to clarify

The object starts from a standstill, accelerates for 2s until reaching full speed which it then does for 4.5s. At this point it immediately reduces speed to a stop over 3s. When it stops it will have travelled a distance of 8509mm. I need to find accel (mm/s^2), full speed (mm/s) and neg accel (mm/s^2)
 
Hi, sorry to clarify

The object starts from a standstill, accelerates for 2s until reaching full speed which it then does for 4.5s. At this point it immediately reduces speed to a stop over 3s. When it stops it will have travelled a distance of 8509mm. I need to find accel (mm/s^2), full speed (mm/s) and neg accel (mm/s^2)
Please check the data for final distance (Is it really given in "mm"?).

If I were to do this problem, I would sketch the approximate v-t curve.

Please show us what you have tried and exactly where you are stuck.​
Please follow the rules of posting in this forum, as enunciated at:​
Please share your work/thoughts about this assignment.​
 
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Hi, sorry to clarify

The object starts from a standstill, accelerates for 2s until reaching full speed which it then does for 4.5s. At this point it immediately reduces speed to a stop over 3s. When it stops it will have travelled a distance of 8509mm. I need to find accel (mm/s^2), full speed (mm/s) and neg accel (mm/s^2)
what about the equations that connect position with time and acceleration?
 
An ASCII art sketch to help you out...
Code:
Veclocity(mm/s)
     |
Vmax |      -------------------
     |     /                   \
     |    /                      \
     |   /                        \
     |  /                           \
     | /                             \
     |/                                \
  0  +-----------------------------------------> time(s)
    0.0     2.0               6.5       9.5
What does the area under the curve represent?
 
Hi

Basically this data has come from a list which shows how an automated scenery truck should move. The guys in the US have basically given me an acceleration time and acceleration (or neg accel) time and a time at full speed they also have told me how far the piece has travelled. I have converted the imperial data into metric which gives the travel distance in this example as 8509mm.

Our system uses acceleration and speed in mm/s^2 and mm/s respectively and therefore I need to find a way of converting the US times into accelerations and speeds. In a separate document they have mentioned that in this example the full speed would be 1,215.644 mm/s so based on that I can assume the acceleration is v/t so 607.822 mm/s^2 and the neg accel is 405.214 mm/s^2.

The trouble is we do not have this full speed number quoted in the 2nd document for every move and therefore I need to derive this full speed from what I was given (accel time, full speed time, decel (neg accel) time and total distance).

Thanks of your time and effort so far - I feel like you are trying to point out I'm missing something but my head is a scrambled mess so I cant see the wood for the trees
 
An ASCII art sketch to help you out...
Code:
Veclocity(mm/s)
     |
Vmax |      -------------------
     |     /                   \
     |    /                      \
     |   /                        \
     |  /                           \
     | /                             \
     |/                                \
  0  +-----------------------------------------> time(s)
    0.0     2.0               6.5       9.5
What does the area under the curve represent?
Distance travelled
 
Distance travelled

Correct! Therefore if you can write an expression for the whole area under the curve, you know this will equal 8509mm.

I'll start you off with the area under the first 2 seconds which is...

[math] \frac{1}{2} \times V_{max} \times 2 [/math]
Now work out the area under the curve for time 2 to 6.5, then 6.5 to 9.5 seconds, add them together (also add the above to get total distance) and then write "=8509". This should enable you to calculate \(V_{max}\). Please post back with your work so that we can check it.
 
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So basically i wound up taking:

(x = travel distance, vf = full speed, tf = time at full speed, t1 = acceleration time, t2 = neg accel time, a1 = accel speed and a2 = neg accel speed)


x = vf((0.5*t1) + tf + (0.5*t2))

and rearranged to:

vf = x / ((0.5*t1) + tftf + (0.5*t2))

Using that I was then able to use the following to find:

a1 = vf / t1

and

a2 = vf / t2
 
a is the acceleration in meters per sq sec

m is the maximum velocity in meters per sec

m = 0 + 2a = 2a

p is the distance traveled in meters while accelerating

p = a(2^2)/2 = 2a

q = distance traveled at maximum velocity in meters per sec

q = 4.5m = 4.5(2a) = 9a

r = distance traveled in meters while decelerating

d = deceleration in meters per sq. sec.

p + q + r = 8.509.

[MATH]m + 3d = 0 \implies d = - \dfrac{m}{3} = \dfrac{2a}{3}\\ \therefore r = 0 + 3^2 * \dfrac{2a}{3} * \dfrac{1}{2} = 3a \implies \\ 2a + 9a + 3a = 8.509 \implies 14a = 8.509 \implies a = \dfrac{8.509}{14} \implies \\ m = 2a = \dfrac{8.509}{7} \approx 1.21557.[/MATH]Your answer looks good to me. Notice that I solved everything in terms of a, which means that you can get approximations for all aspects of this machine's behavior. Next time that you are stuck, define your terms. That will at least organize your thoughts and speed communication if you need to ask for help.
 
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