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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

RMC
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    Calculate the acceleration by taking the second derivative of position in respect to time and then solve for a(t) = 0. Show us your work because most likely you made an algebra error. – Piotr Benedysiuk Jan 27 '16 at 10:01
  • Computing the second derivative I get $$ x''(t) = \frac{12t(4t^2-9)}{(3+4t^2)^3} $$ so that $x''(t)=0$ has solutions $t=0$ and $t=\pm\frac32$. Since $t\geq 0$ you are left with $t=0$ and $t=\frac32$. – AndreasT Jan 27 '16 at 10:06

4 Answers4

1

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.

Mary Star
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First derivative is $\frac{(18+24t^2)-(48t^2)}{(6+8t^2)^2}$ so second derivative becomes $\frac{(6+8t^2)^2.(-48t)-((18-24t^2)(2(6+8t^2)(16t)}{(6+8t^2)^4}=0$ you have an exprssion you will get $t$

Archis Welankar
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If position is given by $x(t)= \frac{3t}{6+8t^2}$, then aceleration, in the classic sense, is $x^{\prime\prime}(t)=\frac{12t(4t^2-9)}{(4t^2-3)^3}$ for all $t\ge0$.

At which aceleration is 0, is $x^{\prime\prime}(t_0)=0$, which explicitely is $t_0=0$, $t_0=3/2$ and $t_0=-3/2$.

Now, since $-3/2<0$, we forgot this solution and get only with $t_0=0$ and $t_0=3/2$. But $t_0=0$ indeffined us the deffiniont of aceleration. Thus $t_0=3/2$.

sinbadh
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$$x(t)=\frac{3t}{6+8t^2}\to x''(t)=a(t)=\frac{\text{d}^2}{\text{d}t^2}\left[\frac{3t}{6+8t^2}\right]$$

So:

$$\frac{\text{d}^2}{\text{d}t^2}\left[\frac{3t}{6+8t^2}\right]=$$ $$\frac{\text{d}}{\text{d}t}\left[\frac{\text{d}}{\text{d}t}\left[\frac{3t}{6+8t^2}\right]\right]=$$ $$\frac{\text{d}}{\text{d}t}\left[3\frac{\text{d}}{\text{d}t}\left[\frac{t}{6+8t^2}\right]\right]=$$ $$3\left[\frac{\text{d}}{\text{d}t}\left[\frac{6-8t^2}{(6+8t^2)^2}\right]\right]=$$ $$3\left[-\frac{32t(6-8t^2)}{(6+8t^2)^3}-\frac{16t}{(6+8t^2)^2}\right]=$$ $$\frac{12t(4t^2-9)}{(4t^2+3)^3}$$

Now solve:

$$\frac{12t(4t^2-9)}{(4t^2+3)^3}=0\Longleftrightarrow$$ $$12t(4t^2-9)=0\Longleftrightarrow$$ $$t(4t^2-9)=0$$

We got tho equations:

  • $$t=0$$
  • $$4t^2-9=0\Longleftrightarrow$$ $$4t^2=9\Longleftrightarrow$$ $$t^2=\frac{9}{4}\Longleftrightarrow$$ $$t=\sqrt{\frac{9}{4}}\Longleftrightarrow$$ $$t=\frac{3}{2}$$
Jan Eerland
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