# Higher Derivatives - Acceleration help

## Homework Statement

A particle moves along the x-axis, its position at time t is given by
$$\displaystyle \Large x(t)=\frac{7 t}{6+5 t^2}, \ t\geq 0,$$
where t is measured in seconds and x is in meters. The acceleration of the particle equals 0 at time t = ____ and ____ seconds.

## Homework Equations

$$\displaystyle \Large x(t)=\frac{7 t}{6+5 t^2}, \ t\geq 0,$$
v(t) = x'(t)
a(t)=v'(t)=x''(t)

## The Attempt at a Solution

once i find a(t), which is the 2nd derivative of the original function, i get:
$$\displaystyle \Large a(t)=\frac{7t-7}{2t(6+5t^2)}$$

so when acceleration=0, $$\displaystyle \Large a(t)=\frac{7t-7}{2t(6+5t^2)}=0$$
t=1
but what the other value?

You solved the derivatives incorrectly

You solved the derivatives incorrectly

oh my bad. it should be:
-440t / (6+5t^2)

now what?

Last edited:
someone?

You still do not have the correct second derivative. (I think).

ok.. i give up. my algebra is good but i have no idea where to go.
this is what i got for the second derivative after doing it 20 times to make sure i didnt make mistakes:
$$\displaystyle \Large x''(t)=\frac{ -70t(6+5t^2)-20t(42-35t^2) }{(6+5t^2)^2} \$$$$\displaystyle \Large =\frac{ -1260t-1050t^3 }{(6+5t^2)^2} \$$

maybe im stuck on the factoring of (42-35t^2)... but i just dont know aaaah

If that is the correct second derivative - and I have not checked - you are now looking for the points at which acceleration equals zero. So you set the second derivative equal to zero and solve for t.

2nd derivative is still wrong
$$\frac{dx}{dt}=-7\frac{5t^{2}-6}{(5t^{2}+6)^{2}}$$
$$\frac{d^{2}x}{dt^{2}}=-7\frac{10t(5t^{2}+6)^{2}-20t(5t^{2}+6)(5t^{2}-6)}{(5t^{2}+6)^{4}}=-70t\frac{(5t^{2}+6)-2(5t^{2}-6)}{(5t^{2}+6)^{3}}=-70t\frac{5t^{2}-18}{(5t^{2}+6)^{3}}$$

2nd derivative is still wrong
$$\frac{dx}{dt}=-7\frac{5t^{2}-6}{(5t^{2}+6)^{2}}$$
$$\frac{d^{2}x}{dt^{2}}=-7\frac{10t(5t^{2}+6)^{2}-20t(5t^{2}+6)(5t^{2}-6)}{(5t^{2}+6)^{4}}=-70t\frac{(5t^{2}+6)-2(5t^{2}-6)}{(5t^{2}+6)^{3}}=-70t\frac{5t^{2}-18}{(5t^{2}+6)^{3}}$$

oh so the only thing i did wrong is including the 7 in my calculations all the time.. hh tnx man!

David Gould - yeah i know that but the thing is i got 0 all the time and i needed another value so here i got it.. tnx