How to find velocity and acceleration at a point given acceleration in terms of position [closed]

Question

A particle moves along a horizontal straight line with an acceleration $𝑎=6 𝑠^{1/3} m/s^{2}$. When $t = 2 s$, its displacement $S = 27 m$ and its velocity $v = 27 m/s$. Calculate the velocity and acceleration of the point when t = 4 s.

My first guess was to integrate with respect to time in order to get the velocity equation. However, this yields an equation that requires to know the position. If I integrate again, the position equation will also require position/displacement. Thus I am achieving nothing by doing so.

My professor gave us the following (among other equations):

$a=f(s)$

Hence, $a ds = v dv$

=> $v dv= f(s) dv$

How can I approach this problem? I know I have to get time as a variable somehow, but I do not know how.

Edit: I do not know if I have to use the previous identity. It's simply my next guess.

Answer 1

When you have acceleration as a function of displacement you can use conservation laws to describe the velocity-position relationship. Do the energy after minus energy before = work done. Work is $W=\int F \,{\rm d}s$ and thus

$$ \tfrac{1}{2} m v^2 - \tfrac{1}{2} m v_1^2 = \int \limits_{s_1}^s (m a)\,{\rm d}s $$

with $s_1 = 27\text{ [m]}$ and $v_1=27\text{ [m/s]}$ and $a = 6 s^\tfrac{1}{3} \mathrm{ [m/s^2]}$. You can cancel out $m$ and solve for $v(s)$.

The solve for time-diplacement with

$$ t = t_1 + \int \limits_{s_1}^s \tfrac{1}{ v(s)} \,{\rm d}s $$

with $t_1=2\text{ [s]}$.

Answer 2

$$ a=f(s)=6\,s^{1/3}=\frac{d^2s}{dt^2}$$

solve this differential equation with the initial conditions $~s(t=2)=27\,,D(s)(t=2)=27~$ you obtain $s(t)$

$$s(t)=(t+1)^3$$

and

$$v=\frac{ds}{dt}=3\,(t+1)^2$$

thus:

the acceleration a

$$a(t=4)=6\,s(t=4)^{1/3}=30~ \text{[m/s^2]}$$

$$v(t=4)=75 ~\text{[m/s]}$$