Find position and velocity at specific time

[JJBladester]

Homework Statement

The acceleration of point A is a = -5.4sin(kt) ft/s².

k = 3 rad/s.

When t = 0, x = 0 and v = 1.8 ft/s.

Determine the position and velocity of point A when t = 0.5s

Answers: x = 0.598 ft v = 0.1273 ft/s

Homework Equations

[URL]http://latex.codecogs.com/gif.latex?x=distance[/URL]

[URL]http://latex.codecogs.com/gif.latex?\dot{x}=velocity[/URL]

[URL]http://latex.codecogs.com/gif.latex?\ddot{x}=acceleration[/URL]

[PLAIN]http://latex.codecogs.com/gif.latex?\dot{x}(t_{2})=\dot{x}(t_{1})+\int_{t_{1}}^{t_{2}}\ddot{x}dt

[PLAIN]http://latex.codecogs.com/gif.latex?x(t_{2})=x(t_{1})+\int_{t_{1}}^{t_{2}}\dot{x}dt

The Attempt at a Solution

The velocity of point A at t = 0.5s is:

[PLAIN]http://latex.codecogs.com/gif.latex?\dot{x}(t_{2})=\dot{x}(t_{1})+\int_{t_{1}}^{t_{2}}\ddot{x}dt

[URL]http://latex.codecogs.com/gif.latex?\dot{x}(0.5)=1.8+\int_{0}^{0.5}[-5.4sin(kt)]dt=0.1273ft/s[/URL]


To find the position of point A at t = 0.5s, I want to use this formula
[PLAIN]http://latex.codecogs.com/gif.latex?x(t_{2})=x(t_{1})+\int_{t_{1}}^{t_{2}}\dot{x}dt

But, since I don't have a formula for [tex]\dot{x}[/tex], I'm not sure how to go about solving for the position of point A at time t = 0.5s.

So far, I have:

[URL]http://latex.codecogs.com/gif.latex?x(t_{2})=\int_{0}^{0.5}\dot{x}dt[/URL]

 

[kuruman]

Can you find a functional form for the velocity v(t) first? Use

[tex]v(t)=\int^{t}_{0}a(t)dt[/tex]

Once you have that, you can evaluate at any time. For the position, don't forget that x(t) is proportional to a(t). What is the constant of proportionality?

 

[JJBladester]

So, if

and

[URL]http://latex.codecogs.com/gif.latex?\ddot{x}=-5.4sin(kt)[/URL]

then

[URL]http://latex.codecogs.com/gif.latex?\dot{x}=\int_{0}^{t}[-5.4sin(kt)]dt[/URL]

What is the constant of proportionality?

I know that:

v=d/t

a=v/t=d/t²

so

d = a*t²

The ratio of distance to acceleration is then

d/a = t²

Is that in the right direction for what you're asking? Now how does that "contstant of proportionality" help me?

 

[kuruman]

So, if

and

[URL]http://latex.codecogs.com/gif.latex?\ddot{x}=-5.4sin(kt)[/URL]

then

[URL]http://latex.codecogs.com/gif.latex?\dot{x}=\int_{0}^{t}[-5.4sin(kt)]dt

[/URL]
Can you do this integral?

I know that:

v=d/t

You don't know that. This expression works only if the acceleration is zero - not the case here.

a=v/t=d/t²

Not true for reasons explained above.

If "constant of proportionality" doesn't ring a bell, integrate v(t) once more to get x(t), then evaluate at the appropriate times. Be sure your expressions reduce to the correct values at t = 0.

 

[JJBladester]

Okay, I have the answers now, but why are we not adding a constant of integration when integrating up from acceleration to velocity to position? Like a v₀ and x₀ term?

__________________________
My solution
__________________________

a(t) = -5.4sin(kt)
v(t) = int(a(t)) = (5.4/k)cos(kt)
x(t) = int(v(t)) = (5.4/k²)sin(kt)

At 0.5s, v(t) = (5.4/3)cos(3*0.5) = 0.1273
At 0.5s, x(t) = (5.4/3²)sin(3*0.5) = 0.598

So I got the answers which kind of makes me happy, but why doesn't the v(t) equation have a v₀ term (constant of integration)? Likewise, why doesn't the x(t) equation have something like v₀t + x₀

By the way, thanks for your help... And I appreciate you not just giving me the answers flat-out... Can't learn that way!

 

[kuruman]

So I got the answers which kind of makes me happy, but why doesn't the v(t) equation have a v₀ term (constant of integration)? Likewise, why doesn't the x(t) equation have something like v₀t + x₀

You add a constant of integration when you do math not physics. In physics there are always limits of integration that take into account the needed constants. More formally, you start from dv = a(t)dt and integrate with limits of integration as shown below

[tex]\int^{v}_{v_0}dv=\int^{t}_{0}a(t)dt[/tex]

Note that this describes physical reality and says that "at t = 0 the velocity is v₀ and at t = t, it is v" which basically establishes v(t) and takes into account v(0). Do it formally this way and see what you get.

 

[JJBladester]

You add a constant of integration when you do math not physics. In physics there are always limits of integration that take into account the needed constants. More formally, you start from dv = a(t)dt and integrate with limits of integration as shown below

[tex]\int^{v}_{v_0}dv=\int^{t}_{0}a(t)dt[/tex]

Note that this describes physical reality and says that "at t = 0 the velocity is v₀ and at t = t, it is v" which basically establishes v(t) and takes into account v(0). Do it formally this way and see what you get.

Your explanation was fantastic. Here is what I've got.

 

[kuruman]

Your explanation was fantastic. Here is what I've got.

Great. Now you are ready to understand the "constant of proportionality." Note that

a(t) = -5.4 sin(kt)
x(t) = (5.4/k²) sin(kt)

In other words, x(t) = -(1/ k²) a(t), which says that the position at any time t is proportional to the acceleration at any time t. That's what I had in mind when I posted my first response to your question.

 

[JJBladester]

In other words, x(t) = -(1/ k²) a(t), which says that the position at any time t is proportional to the acceleration at any time t.

Ok, so in this example, the constant of proportionality states a relationship between position and acceleration. Knowing that x(t) is -1/k²a(t), I can just find the acceleration at any point in time and multiply it by -1/k² to find the position at that time. Neat.

 

[kuruman]

And once you know x(t), you take its time derivative to find v(t). It's a shortcut.