derivatives
A particle moves along the x-axis, its position at time t is given by
$x(t)= \frac{3t}{6+8t^2}$, $t≥0$,
where t is measured in seconds and x is in meters.
Find time at which acceleration equals 0.
I got the answer 0.5 and 0 but apparently i got the answer wrong. I know I need to use derivative twice. Can someone please help me
Calculation
$velocity = \frac {-3(4t^2-3)}{2(4t^2+3)^2}$
$Acceleration = 36x(16x^4+8x^2-3)$
answer 0.5 and 0 was found through wolfram
The best you can do in this case is find the average velocity.
The average velocity, $v_{\text{avg}}$ is given by: $$ v_{\text{avg}}=\frac{\Delta s}{\Delta t} $$ Where $\Delta s$ is the change in displacement and $\Delta t$ is the change in time. ($\Delta$ is the Greek letter "delta" and in this context, it is the final value take away the initial value, so for example $\Delta t=t_{\text{final}}-t_{\text{initial}}$)
At we want the velocity at $t=3$, so we can use the two points around it ($t=2.7\ \text{s}$ and $t=3.2\ \text{s}$): $$ v_{avg}=\frac{\Delta s}{\Delta t}=\frac{10.6-7.8}{3.2-2.7}=5.6\ \text{m}/\text{s} $$
Velocity is a vector quantity, and indicates both speed (by its slope) and direction (by its sign). Speed is a scalar quantity, and represents, colloquially, how "fast" the particle is moving (distance over time). And because it doesn't matter in which direction the particle is moving, speed is represented by $|v|$. As Spencer commented, when velocity and acceleration are both positive or both negative, the speed is increasing. When they are different signs, then the speed is decreasing.
To see why, look at this portion of the graph of $x^3$ as x approaches 0. The particle's graph is going up for sure (positive velocity). However, the rate by which its increasing is decreasing (negative acceleration) -- hence why its increasing ever more gradually. In other terms, it's slowing down, because negative acceleration indicates a decreasing velocity.
The same would apply to the converse as well -- a positive acceleration and a negative velocity would mean a graph which is decreasing ever more slowly, because the velocity is still negative but is increasing. So the "graduating effects" are caused by opposite signs.
Conversely, when velocity and acceleration have the same sign, the graph either "shoots up" or "shoots down", which is what you are seeing here:
Either the velocity is increasing increasingly, or it is decreasing increasingly, depending on whether the signs are both positive or negative (this is a graph of positive signs).
The key point is that speed doesn't have regard for direction -- so whether or not the particle is going left or right is not of your concern. Hence the $|v|
Best Answer
The acceleration is expressed by the second derivative.
We have $$x(t)=\frac{3t}{6+8t^2}$$ $$x'(t)=\frac{3(6+8t^2)-3t\cdot 16t}{(6+8t^2)^2}=\frac{18+24t^2-48t^2}{(6+8t^2)^2}=\frac{18-24t^2}{(6+8t^2)^2}$$ $$ x''(t)=\frac{-48t(6+8t^2)^2-2(18-24t^2)(6+8t^2)(16t)}{(6+8t^2)^4}=\frac{(6+8t^2) \left [-48t(6+8t^2)-2(18-24t^2)(16t)\right ]}{(6+8t^2)^4}=\frac{ 16\left [-3t(6+8t^2)-2t(18-24t^2)\right ]}{(2(3+4t^2))^3}=\frac{ 16\left [-6t(3+4t^2)-6t(6-8t^2)\right ]}{8(3+4t^2)^3}=\frac{-12t\left [3+4t^2+6-8t^2\right ] }{(3+4t^2)^3}=\frac{-12t\left [9-4t^2\right ] }{(3+4t^2)^3}$$
Therefore, the acceleration of the particle equals $0$ at time $t=0$ seconds and $t=\frac{3}{2}$ seconds.