When is the velocity of a particle zero while acceleration is also zero?

[DontEvenTry]

Homework Statement

The motion of a particle is described by the position function

s=((1/3)t^3)-(29t) - (1296/t), t>=1,
where t is the time in seconds and s is the displacement in metres

a) What is the velocity when acceleration is zero?
b) Find the displacement when v = 68m/s
c) When is the acceleration -90m/s^2?

Homework Equations

The Attempt at a Solution

a) Well, if the velocity is ((t^2)-(29)+(1296t^-2), and the acceleration is (2t^4 - 2592)/t^3, then wouldn't the velocity be 0 when acceleration is 0?

b) v(68)= 4595

c)-90 = (2t^4 - 2592)/t^3 ? (Doesnt seem to work)

I know I am doing something terribly wrong

 

[gabbagabbahey]

Hi DontEvenTry, welcome to PF!

Position, velocity, and acceleration are usually vectors; before we proceed, are you sure the position is given as the scalar function you've written above and not as a vector?

 

[HallsofIvy]

gabbagabbahey, I would interpret this as "one-dimensional" motion. The only distinction between "velocity" and "speed" in one dimension is that velocity can be negative (and this clearly can, for example, v(1) is negative) while speed is always non-negative.

Homework Statement

The motion of a particle is described by the position function

s=((1/3)t^3)-(29t) - (1296/t), t>=1,
where t is the time in seconds and s is the displacement in metres

a) What is the velocity when acceleration is zero?
b) Find the displacement when v = 68m/s
c) When is the acceleration -90m/s^2?

Homework Equations

The Attempt at a Solution

a) Well, if the velocity is ((t^2)-(29)+(1296t^-2), and the acceleration is (2t^4 - 2592)/t^3, then wouldn't the velocity be 0 when acceleration is 0?

Why would you think that? Surely you don't think "acceleration 0" means "sitting still". A car going at a constant speed of 90 mph has "acceleration 0". You are correct that the acceleration is (2t^4- 2592)/t^3. Now solve (2t^4- 2592)/t^3= 0 and put that value of t into the formula for v.

b) v(68)= 4595

? The problem did not ask for v, it asked for "displacement"- the distance moved. Since v= dx/dt, $x= \int v dt$.

c)-90 = (2t^4 - 2592)/t^3 ? (Doesnt seem to work)

Why not? This is the same as -90t^3= 2t^4- 2592 or 2t^4+ 90t^3- 2592. Solve that. (Hint: it has a simple integer solution.)

I know I am doing something terribly wrong

Yes, you are giving up too easily- that's terribly wrong!
How do I know there is a "simple integer root"? I tried. Since cubics and quartics can be horrible to solve in general, I guessed (hoped) that there was a simple integer root and just tried putting integer values into the formula.

 

[DontEvenTry]

Alright, for a) i got 71m/s
b) 272m
c) 3sec