How to Calculate Velocity with Given Acceleration and Position Using Integration

[Phys_Boi]

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?

 

[berkeman]

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) = ?

 

[Phys_Boi]

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)?

 

[berkeman]

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?

 

[Phys_Boi]

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?

 

[berkeman]

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);

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

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

 

[Phys_Boi]

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.

 

[berkeman]

Your welcome. They are not ignorant questions -- you are learning.

 

[berkeman]

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.

 

[Phys_Boi]

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!