## Physics Particle Movement Calculations

1. The problem statement, all variables and given/known data
A particle moves in such a way that its position z, in meters, is given as a function of time t, by the equation z = 2t^2 − 3t^3. At what times is the particle at position z = 0? (two answers; one is 6E-1)

2. Relevant equations
z = 2t^2 − 3t^3

3. The attempt at a solution
I've tried plugging in certain numbers and i'm trying to figure out what my teacher means by 6E-1. Refreshing my memory of calculus etc..

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 Recognitions: Homework Help 6E-1 means 6 x 10 to the power -1, which is 0.6. You can put 6E-1 into your calculator, press ENTER or = and see that. I don't think 6E-1 is a correct solution of that equation. Better check the question again. You have a cubic equation to solve, luckily an easy one. The usual approach is to factor it. Look for a common factor. For example, if you had this equation: 0 = 3x - 9, you would think as follows: The factors of 3x are 3 and x. The factors of the second term, 9, are 3 and 3. The factor in both is 3. So you would write down the 3, put brackets and ask yourself what 3 must be multiplied by to get the original 3x - 9: 0 = 3x - 9 = 3(x - 3) To solve it, you ask if each factor could be zero. 3 can't be zero, but x-3 can. x - 3 = 0 x = 3 (from adding 3 to both sides) example 2: 0 = 2t^3 + t^2 - t t is a common factor of all 3 terms. So 0 = t(2t^2 + t - 1) A method called "trinomial factoring" can be used on the factor in the brackets. It is a bit too complicated to write easily in plain text, so I'll leave you to look it up. 0 = t(t+1)(2t-1) Then t = 0 or t+1 = 0 or 2t-1 = 0 and the 3 solutions are t = 0, t = -1 and t = 1/2
 Thank you very much for your help! I came out with the following conclusion: factored z=2t^2-3t^3 into z=t^2(2-3t) so I concluded that one answer is t=0 because 0^2(anything) would = 0 I also thought 2-3(.666666forever don't know how to write that on comp) would also equal 0 because 3*0.6forever would equal 2 so 0.6 was close. Is my thinking right? Thanks again for your help very much appreciated.

## Physics Particle Movement Calculations

Ok, new question related to this one:

Use Diﬀerentiation to ﬁnd an expression for the velocity of the particle as a function of time.

I can't figure out what Differentiation is any ideas

Thanks

 Quote by Jordash Thank you very much for your help! I also thought 2-3(.666666forever don't know how to write that on comp) would also equal 0 because 3*0.6forever would equal 2 so 0.6 was close. Is my thinking right? Thanks again for your help very much appreciated.

If your function was 2t^2-3t^3 you can get t^2(2-3t)=0, so t=o and then:

2-3t=0 ---> -3t=-2 ---> t=(2/3) = .6666 (your teachers answer of 6E-1 i guess).

As far as differentiation, when you differentiate a position function, you get a velocity function. When you differentiate a velocity function

Im assuming you didnt learn differentiation yet? All you have to know for this question is the power rule:

say you have ax^n, to differentiate put it in the form of nax^n-1...

So heres an example:

7x^5 ---->after differentiation you get 35x^4

 Tags calculation, movement, particle, physics, velocity