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!!
Best Answer
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.