Question

Mechanics Problem:

Answer 1

11.1 The motion of a particle is defined by the relation \[x=1.5t^4-30t^2+5t+10\], where x and t are expressed in meters and secods, respectively. Determine the position, the velocity, and the acceleration of the particle when t = 4 s

Answer 2

Does the coefficients has units? If so, what does these units represents?

Answer 3

Does \[\frac{\text{m}}{\text{s}^4}\], \[\frac{\text{m}}{\text{s}^2}\], \[\frac{\text{m}}{\text{s}^2}\]

Answer 4

mean something in the physical sense?

Answer 5

I graphed the function, and I go a weird movement:

Answer 6

dude differentiate x wrt to t.it will dive you velocitydx/dt=v
double differentiate to get accn

Answer 7

yeah simply diff once to get velocity at any time, and twice to get acceleration.

Answer 8

Thank you fortheloveofscience, but I don't want to obtain neither acceleration nor velocity. I was just wondering about the physical meaning of the units involved. I mean I don't know what does it mean m/s^4

Answer 9

ohkay i got it.nope they are the parametric terms.^4 only means a value

Answer 10

I mean by looking at the units I know that -30 is an acceleration and 5 a velocity. But I don't what 1.5 is.

Answer 11

not such as that.u have to first diff. it first

Answer 12

Well I think I don't need to diff to find the acceleration or initial velocity.

Answer 13

Start with the derivation of the position of a particle with a uniform acceleration. This is done by repeated integration with respect to time:\[a=a\]\[v=at+v_i\]\[x=\frac12at^2+v_it+x_i\]Every constant that we pick up from integration will be the initial value of the quantity we get as a result. For example the constant we get from integrating adt is the initial velocity.
So in our formula you can see that the coefficients of the terms are 1/2 acceleration, the initial velocity, and the initial position. The physical significance of the units m/s^2 on the first coefficient represent the fact that it is a change per second of velocity, which is in m/s (a=m/s/s=m/s^2)
Now note that the derivative of acceleration with respect to time is a quantity called 'jerk' (so named because it feels like a jerk when you experience it). Jerk will of course have units of m/s^3. Relating this to the physical significance of m/s^2 for acceleration, these units should not be too confusing.
If we do the same derivation for a situation with constant jerk we get\[J=J\]\[a=Jt+a_i\]\[v=\frac12Jt^2+a_it+v_i\]\[x=\frac16Jt^3+\frac12a_it^2+v_it+9x_i\]Now what if the jerk is not constant? I don't know if that quantity has a name, but let's call it Q. If Q is a constant (that is, the jerk is changing linearly) we can derive\[Q=Q\]\[J=Qt+J_i\]\[a=\frac12Qt^2+J_it+a_i\]\[v=\frac16Qt^3+\frac12J_it^2+a_it+v_i\]\[x=\frac1{24}Qt^4+\frac16J_it^3+\frac12a_it^2+v_it+x_i\]So the first coefficient is 1/24 the change in jerk with respect to time, the second is 1/6 the change in acceleration with respect tot time, etc...

Answer 14

Ohh that is what I wanted, Thank you so much TT!

Answer 15

anytime!
it's interesting question and I had to think about it.
*small typo above: that 9 next to the x in the second part does not belong