Equations of Motion

**zoolander82** · Jun25-09, 02:08 PM

It's been at least 5 or 6 years since I've done any form of calculus so I'm very rusty :( I'm trying to determine the time it takes to achieve constant velocity of a motor I'm trying to spec out/replace.

Here's what I've got: We have a motor initially at rest which is driving a platform and then comes to rest again at 6.3s. I'm considering linear motion here. Once the motor is turned on, I assumed the acceleration to be linear until it reaches constant acceleration (and eventually constant velocity). I know this is not a realistic assumption however, it gives me a good idea of values.

This gives rise to 3 sections on a graph and 4 time periods. t0 = 0s, t1, t2 and t3 = 6.3s. 0<t<t1 accel. is linearly increasing, t1<t<t2 is constant accel. and t>t2 accel. = 0. My additional values are as follows: constant acceleration is 45mm/s^2 and constant velocity is 6mm/s. In other words:

a(0) = v(0) = 0
a(t1) = 45 mm/s^2
a(t2) = a(t3) = 0
v(t2) = v(t3) = 6 mm/s

I've tried solving this in the 3 segments (as outlined above).
0<t<t1:
a(t) = (45/t1)*t;
v(t) = (45/2*t1)*t^2 + C;
evaluating the definite a(t) integral from 0 -> t1, v(t) = (45*t1)/2

I want to solve for t1...this is where I get lost. I say that t1 must equal t in order for me to get 45. That doesn't make sense though because t can be any real number which doesn't help me...moving forward.

t1<t<t2:
a(t) = 45;
v(t) = 45t + C;
evaluating the definite a(t) integral from t1 -> t2, v(t) = 45*(t2-t1)

Now I want to solve for t2. I should know t1 from the above equation. If I say t1 = 1s, then t2 = 1.133s. This seems to make sense to me, but I can't be 100% certain. Any help is appreciated.

**HallsofIvy** · Jun25-09, 02:47 PM

Originally Posted by zoolander82

It's been at least 5 or 6 years since I've done any form of calculus so I'm very rusty :( I'm trying to determine the time it takes to achieve constant velocity of a motor I'm trying to spec out/replace.

Here's what I've got: We have a motor initially at rest which is driving a platform and then comes to rest again at 6.3s. I'm considering linear motion here. Once the motor is turned on, I assumed the acceleration to be linear until it reaches constant acceleration (and eventually constant velocity). I know this is not a realistic assumption however, it gives me a good idea of values.

This gives rise to 3 sections on a graph and 4 time periods. t0 = 0s, t1, t2 and t3 = 6.3s. 0<t<t1 accel. is linearly increasing, t1<t<t2 is constant accel. and t>t2 accel. = 0. My additional values are as follows: constant acceleration is 45mm/s^2 and constant velocity is 6mm/s. In other words:

a(0) = v(0) = 0
a(t1) = 45 mm/s^2
a(t2) = a(t3) = 0
v(t2) = v(t3) = 6 mm/s

I've tried solving this in the 3 segments (as outlined above).
0<t<t1:
a(t) = (45/t1)*t;
v(t) = (45/2*t1)*t^2 + C;

And, since v(0)= C= 0, v(t)= (45/(2t1))t^2.

evaluating the definite a(t) integral from 0 -> t1, v(t) = (45*t1)/2

You mean v(t1)= (45t1)/2. And, since you got v(t) by integrating a(t) that is the same as putting t= t1 in your formula for v(t)

I want to solve for t1...this is where I get lost. I say that t1 must equal t in order for me to get 45. That doesn't make sense though because t can be any real number which doesn't help me

You can't find t1 from this information. You want to get a final acceleration of 45 but you have defined the rate or change of acceleration itself in terms of t1. You must be given either the rate of change of acceleration or t1 itself.

....moving forward.

t1<t<t2:
a(t) = 45;
v(t) = 45t + C;
evaluating the definite a(t) integral from t1 -> t2, v(t) = 45*(t2-t1)

Now I want to solve for t2. I should know t1 from the above equation. If I say t1 = 1s, then t2 = 1.133s. This seems to make sense to me, but I can't be 100% certain. Any help is appreciated.

**zoolander82** · Jun25-09, 03:42 PM

Sorry, I just realized I left out a pretty important piece of information. The total displacement is 32.8mm. So now, basically I should be able to guess and check with t1 and t2 until my displacement equals 32.8, right? There is a function in Excel that does this, but I can't remember what it is. Anyone know what the function name is and where to find it in Office 2007?