Problem: A particle moves in a straight line with position relative to the origin given by $s(t)= 2t-\frac{4}{t+1}$ cm, where $t ≥ 0$ is the time in seconds.

Here's what I know:

a) Find velocity and acceleration functions for the particles motion.

  • $v(t)= 2+\frac{4}{(t+1)^2}$
  • $a(t)= \frac{-8}{(t+1)^3}$

b) Describe the motion of the particle at $t= 1$ second.

  • $s(1)= 0 cm$
  • $v(1)= 3 cm/s$
  • $a(1)= -1 cm/s^{2}$

Therefore, the particle is located at point $O$ and moving to the right at a speed of $3 cm/s$. Since $a$ and $v$ have different signs, the speed of the particle is decreasing.

c) Does the particle ever change direction? If so, where and when does it do this?

The particle has to be stopped to change direction, so I try and find where the particle stops.

I set $v(t)= 0$.

After attempting to solve for $v(t)=0$ , I discover the answer does not exist because ${(t+1)^2}=-2$ cannot be solved since you cannot find the square root of a negative number.

Here's what I'm stumped on:

d) Find the time intervals when the velocity is increasing.

Please and thank you!!

  • $\begingroup$ Hint: If a partile is changing direction, what can you say about its velocity just before it changes direction? For your second question, this is simply finding $t$ for when $v(t) > 0$. $\endgroup$ Commented Nov 27, 2017 at 4:36
  • $\begingroup$ I've added some mathjax, to your post, which might help you in the future (math.meta.stackexchange.com/questions/5020/…). That said, excellent first post, and welcome to MSE! $\endgroup$
    – B. Mehta
    Commented Nov 27, 2017 at 4:39
  • $\begingroup$ Hint: Describe the motion of the particle at more than the one time suggested. Say $t= 0, 2, 3, 4$. You might want to put the results in a table or graph them. They should tell you something about the motion. While your calculation of the motion of the particle at $t=1$ is correct, your description of it is not. $\endgroup$ Commented Nov 27, 2017 at 4:51

1 Answer 1


Since velocity is continuous for $t\geq 0$, by the intermediate value theorem, the only points at which $v$ could potentially change sign is when $v=0$. This would require $$ (t+1)^2=-2 $$ which is not possible, as the square of a real number cannot be negative.

As for your second question, checking where $a(t)>0$ will suffice.


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