Speed distance velocity and other headaches
Hi all.
I need to calculate the speed an object needs to move at to get to a certain point over a period of time at a constant acceleration
I have
Vo  initial velocity
Xo  start point
X end point
a  constant acceleration
t  time
So I need to know the velocity if I am going to add a constant acceleration so the object will get from point Xo to X in the set time.
I also need to know how you calculate the following in a simple formula:
(((((1+2)+3)+4)+5)+6)
Thanks
I need to calculate the speed an object needs to move at to get to a certain point over a period of time at a constant acceleration
I have
Vo  initial velocity
Xo  start point
X end point
a  constant acceleration
t  time
So I need to know the velocity if I am going to add a constant acceleration so the object will get from point Xo to X in the set time.
I also need to know how you calculate the following in a simple formula:
(((((1+2)+3)+4)+5)+6)
Thanks
Quote:(((((1+2)+3)+4)+5)+6)
Why can't you just remove the brackets?
Or are you trying to go for something like recursion (better off going from 6 to 1)?
Or a for loop?
(somewhere above)
result = 0
for(x=1;x<7;x++) {
result = result + x
}
Sorry, the brackets were unnecessary. I want to work out how much constant acceleration was added to an object over a period of time
To answer to your initial question,
x = x0 + v0*t + (a*t*t)/2
So if you know t, x, a and x0, you can find v0.
Which would be the initial speed of the object in order to reach x from x0 under acceleration a for time frame t.
If you want to know other quantities remember these relationships:
x = v*t
v = a*t
 Kjurtyl.
x = x0 + v0*t + (a*t*t)/2
So if you know t, x, a and x0, you can find v0.
Which would be the initial speed of the object in order to reach x from x0 under acceleration a for time frame t.
If you want to know other quantities remember these relationships:
x = v*t
v = a*t
 Kjurtyl.
Thanks, I will try it out.
I figured out the other one:
if a = acceleration constant and t = the number of times the acceleration is applied then
((a+(a*t))/2)*t
I figured out the other one:
if a = acceleration constant and t = the number of times the acceleration is applied then
((a+(a*t))/2)*t
The sum of all numbers from 1 to X is (X*(X+1))/2.
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