# Particle moves according to the following equation

1. Feb 23, 2017

### alex91alex91alex

So I had a test and this was the only exercise I miss points at and I can not understand why, I did everything according to my class notes.

A particle moves according to the following equation: x=( −0.00521 cm/s^4 ) t^4 +( 0.0895 cm/s^3 ) t^3 + ( 0.335 cm/s^2 ) t^2−( 1.881 cm/s ) t−1.237 cm

a) What is the initial position of the particle?
x0=-1.237 cm (correct)

b) What is the initial velocity of the particle?
v0= -18.881 cm/s (correct)

c) What is the particle's initial acceleration?
a0= 0.355 cm/s^2 (incorrect), don't we just take the values from the formula given for x0, v0 and a0? That is what I have on my notes.

d) What is the particle's initial jerk?
j0=0 (incorrect) isn't the j0 always 0 regardless of the equation?

e) What is the particle's position at t = 2.67 seconds?
dx/dt=(-0.02084 cm/s^4) t^3 + (0.20685 cm/s^3) t^2 + (0.67 cm/s^2) t - 1.881 cm/s (correct)

x2.67= 1.4 cm (incorrect) I plug 2.67 for t and got 1.425337693, did I get it wrong because I rounded?

f) What is the particle's velocity at t = 2.67 seconds?
dx/dt= (-0.06252 cm/s^4) t^2 + (0.537 cm/s^3) t + 0.67 cm/s^2 (correct)

v2.67= 1.7 cm/s (incorrect) I plug 2.67 for t and got 1.678091172, did i get it wrong because I rounded?

g) What is the particle's acceleration at t = 2.67 seconds?
da/dt= (-0.12504 cm/s^4) t + 0.537 cm/s^3 (correct)

a2.67= 0.2 cm/s^2 (incorrect) I plug 2.67 for t and got 0.2031432. No way I would get counted wrong because of rounding here. What am i doing wrong?

h) What is the particle's jerk at t = 2.67 seconds?
j2.67= -0.12504 cm/s^4 (incorrect) that the only value left if I take the derivative, still got it wrong.

Sorry for the long post, thanks in advance.

2. Feb 23, 2017

### Staff: Mentor

You are given position as a function of time. How do you find the velocity, acceleration, jerk, at any time? (Not sure where you think those initial values appear in the position equation.)

See above.

Why are you using dx/dt? You're finding position, not speed.

3. Feb 23, 2017

### Simon Bridge

I think part (e) illustrates your problem best, so rather than go through each one, I'll just do that one:
... the statement is true, but it is not the correct approach to solve this problem.
No.
You seem to have misunderstood what the equation is trying to tell you.

You are given x(t) ... this is what tells you the position of the particle x at time t.
dx/dt tells you v(t), the velocity of the particle at time t.
dv/dt tells you a(t), and so on.

So the first question asks for the initial position of the particle, which is x(0)... put t=0 in x(t) and see what you get. (You seem to have just been picking the coefficients that have the correct units.)

Part (e) want's the position at t=2.67s, which is x(2.67), so put t=2.67 into x(t) and see what you get.

The second question wants the initial acceleration, which is a(0) = 0.670cm/s^2 ... because that is what you are left with after taking the derivative twice and putting t=0. From that you should be able to work out the jerk - which is da/dt.

Your reasoning about the jerk is incorrect. The jerk is the change in acceleration with time... if the jerk is always zero, then acceleration can never change.

Try again and keep track of the derivatives each time.

Try writing out the general form:
x(t) = A + Bx + Cx^2 + Dx^3 + Ex^4 so x(0) = ?

v(t) = dx/dt = ...? so v(0) = ?
a(t) = dv/dt = ...? so a(0) = ?
j(t) = da/dt = ...? so j(0) = ?
... compare with your equation.

4. Feb 23, 2017

### alex91alex91alex

Oh, I see what you mean. I had my notes wrong. Thanks for the clarification.

5. Feb 23, 2017

### alex91alex91alex

So, since my derivatives are correct I should get:

x0= -1.237 cm
x2.67=-2.43 cm

v0=-1.881 cm/s
v2.67=1.43 cm/s^2

a0=0.67 cm/s^2
a2.67=1.66 cm/s^2

j0=0.537 cm/s^3
j2.67=0.2 cm/s^3

Would the results be correct now?

6. Feb 23, 2017

### Simon Bridge

Looks good ...
off my last post ...
x(0) = A
v(0) = B
a(0) = 2C
j(0) = 6D

In fact, $x(t) = x(0) + v(0)t + \frac{1}{2}a(0) t^2 + \frac{1}{6}j(0) t^3 + \cdots$
... what you are learning is called the MacLaurin series approximation.