Function of particle

Question

The position function of a particle in rectilinear motion is given by s(t) s(t) = t3 - 9t2 + 24t + 1 for t ≥ 0. Find the position and acceleration of the particle at the instant the when the particle reverses direction. Include units in your answer

Answer 1

The position function of a particle in rectilinear motion is given by s(t) s(t) = t3 - 9t2 + 24t + 1 for t ≥ 0. Find the position and acceleration of the particle at the instant the when the particle reverses direction. Include units in your answer

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Reversing direction means changing of velocity sign.

the motion is rectilinear, along a straight lin? that is; so V can be either + or -

 

And, velocity is:

 

V=ds/dt=3t^2-18t+24

 

This is a parabola, crossing "t" axes twice - in two moments in time the particle reverses its direction.:

We have to find the nearest "future" time - the smalllest "zero" for parabola V.

 

Let's see when V turns to zero; it's happening when

 

V=0 i.e.  3t^2-18t+24=0   , divide by 3

 

               t^2-6t+8=0    , solve the quadratic:

 

         solutions are t=2 and t=4

 

It's happening in positions at s(2)  and s(4)

You have to find s(2)=... and s(4)=...

 

 

The acceleration is, a=dV/dT=6t-18

Further,

            a(2)=6*2-18=-6

            a(4)=+6

 

Units: .... The problem doesn't provisde any units here: "s" can be measured in cm, m,mi, ...

and the time- in s, min, hr, ...

In all cases, V has dimention of [distance]/[time]

                  a  - [distance]/[time^2]

 

This must help.

Check the arithmetics, and complete the computations ...

 

Good luck!

 

P.S. my understanding is, problem requires a single solution (... find the position). Then we must consider the expressons for t=2; unless particle starts its motin in the moment  t>2.

Doble check problem conditions ... the only initial state dta mentioned is: t>0. no units, no initial moment provided.