Maximum particle position

  • #1


I don't believe that this problem is difficult, but I am unsure of myself as to the answers.

The position of the particle moving along an x-axis depends on the time according to: x = ct^2 - bt^3 (x in meters and t in seconds)

a) what units must c and b have? Let their numberical values be 3.0 and 2.0, respectively. b) At what time does the particle reach its maximum positive x position?

This is what I did:

a) m - cs^2 - bs^3 --> ms^2/2^2 - ms^3/s^3 = m - m = __ m

so, c = m/s^2 and b = m/s^3

b) x = ct^2 - bt^3

i am imagining the particle to be something like a ball thrown in the air. is this a good idea?

with that in mind, i would imagine the velocity of the ball at the max point would be 0, so:

v = x' = 6.0t - 6.0t^2 = 0 --> t=1 or t = 0

i am ruling out t = 0, because that is the starting time so i believe that the time it reaches it's max position is at 1.0 seconds.

was my process for a and b correct?
  • #2
Yes, the units of b and c must be such that the "sec" cancel, leaving meters. The power on the s must be exactly the power on t
c is in "m/s^2" and b is in "m/s^3". (You do have a typo: you meant
ms^2/t^2 where you wrote "ms^2/2^2".)

Since x= ct^2+ bt^3 so x'= 2ct+ 3bt^2. As this is a physics problem, you can think "at max or min position the speed must be 0". A math major would be more inclined to think "at max or min, the derivative is 0".

Looks to might like you have it exactly right.
  • #3
Hi missrikku,
I think your answer to a) is correct.
As for b): No values of b, c are given in the problem, so you should keep them as unknowns (not plug in values). Your approach of x' = 0 is IMO OK. It will give you 2 answers of t(c,b). Please check out x''(t) for your answers. If the 2nd derivative is negative, then you have found a maximum.

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