How to Calculate Velocity with Given Acceleration and Position Using Integration

Click For Summary
To calculate velocity given acceleration and position, it's essential to understand the full equations of motion under constant acceleration. The equations v(t) = v(0) + a*t and y(t) = y(0) + v(0)*t + ½a*t² are crucial, requiring initial conditions for accurate calculations. The discussion emphasizes the importance of correctly defining the coordinate system, particularly the signs of acceleration, initial position, and velocity. The user seeks clarity on how to derive velocity from height and acceleration, specifically asking if v = at is sufficient. Understanding these principles will aid in solving motion problems effectively.
Phys_Boi
Messages
49
Reaction score
0
If i have an equation: h = ½at2
And:
40 = -4.9t2 → 0 = -4.9t2 - 40

If: v = ∫a

then how do i find the velocity given acceleration of -9.8 m/s2?
 
Physics news on Phys.org
Phys_Boi said:
If i have an equation: h = ½at2
And:
40 = -4.9t2 → 0 = -4.9t2 - 40

If: v = ∫a

then how do i find the velocity given acceleration of -9.8 m/s2?
You are missing some constants in those equations. Are you familiar with the full equations of motion given a constant acceleration a?

v(t) = ?
x(t) = ?
 
berkeman said:
You are missing some constants in those equations. Are you familiar with the full equations of motion given a constant acceleration a?

v(t) = ?
x(t) = ?
Well I am only a Sophomore and haven't had either Calculus nor Physics..
However, v(t) is what i need to know.
What is x(t)?
 
Phys_Boi said:
Well I am only a Sophomore and haven't had either Calculus nor Physics..
However, v(t) is what i need to know.
What is x(t)?
x(t) is the linear distance travelled. It looks like you are wanting z(t) or h(t) height, but it doesn't matter, you still use the same equations.

See this Hyperphysics page for example:

http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#motcon

It lists the equations for v(t) and x(t) or h(t). Although, they have left off one term -- there should be an x(0) term on the righthand side of the equation for x(t).

Does that help?
 
berkeman said:
x(t) is the linear distance travelled. It looks like you are wanting z(t) or h(t) height, but it doesn't matter, you still use the same equations.

See this Hyperphysics page for example:

http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#motcon

It lists the equations for v(t) and x(t) or h(t). Although, they have left off one term -- there should be an x(0) term on the righthand side of the equation for x(t).

Does that help?
Thanks,

So how do i find the velocity at a given time? Do I need all of those variables?
If height is 40:
40=0.5(9.8)(t^2) ... We can solve this for time, however, how do I find the velocity?
Sorry, if I'm missing something that's obvious.

So does v = at?
 
Phys_Boi said:
Thanks,

So how do i find the velocity at a given time? Do I need all of those variables?
If height is 40:
40=0.5(9.8)(t^2) ... We can solve this for time, however, how do I find the velocity?
Sorry, if I'm missing something that's obvious.

So does v = at?

Your problem statement is still not clear to me. We need to know the initial conditions h(0), v(0) in order to do any calculations. To summarize, you will use equations like these (I'm going to use y(t) for the height);

v(t) = v(0) + a*t

y(t) = y(0) + v(0)*t + \frac{1}{2}a*t^2
 
berkeman said:
Your problem statement is still not clear to me. We need to know the initial conditions h(0), v(0) in order to do any calculations. To summarize, you will use equations like these (I'm going to use y(t) for the height);

v(t) = v(0) + a*t

y(t) = y(0) + v(0)*t + \frac{1}{2}a*t^2

Okay... Thank you!

Sorry for the ignorant questions.
 
Your welcome. They are not ignorant questions -- you are learning. :smile:
 
BTW, it's important that you get the signs right for whatever coordinate system you define. So if y(t) is positive up, then the acceleration in the y direction is -9.8m/s^2 as you have already said. The initial position and velocity will have signs as well in the general problems.
 
  • #10
berkeman said:
BTW, it's important that you get the signs right for whatever coordinate system you define. So if y(t) is positive up, then the acceleration in the y direction is -9.8m/s^2 as you have already said. The initial position and velocity will have signs as well in the general problems.
Okay! Thank you!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
18
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 7 ·
Replies
7
Views
1K