position of particle

Question

The position of a particle moving along the $x$ axis depends on the time according to the equation $x = ct^2 - bt^3$, where $x$ is in meters and $t$ in seconds.

What are the units of (a) constant $c$ and (b) constant $b$? Let their numerical values be $3.0$ and $2.0$, respectively. (c) At what time does the particle reach its maximum positive $x$ position? From $t = 0.0 s$ to $t= 4.0 s$, (d) what distance does the particle move and (e) what is its displacement? Find its velocity at times (f) $1.0 s$, (g) $2.0 s$, (h) $3.0 s$, and (i) $4.0 s$. Find its acceleration at times (j) $1.0 s$, (k) $2.0 s$, (l) $3.0 s$, and (m) $4.0 s$.

I don't understand how to approach this.

I think (a) and (b) are $m/s^2$ and $m/s^3$. I think (c) is $0 = 3t^2 - 2t^3$. I don't know what next.

Answer 1

Your solutions to a) and b) are correct. For c) and d), you just need to find maxima and minima. For the rest, just differentiate then, and plug in the values of $t$ at which you are supposed to find the instantaneous velocities and accelerations (for the remaining parts of the question).

To find the maxima/minima of $x(t)$, you merely set the velocity to $0$ and solve for $t$. So in your particular case, we have $v(t)=\frac{d}{dt}(3t^2-2t^3)=6t-6t^2=0\rightarrow t(1-t)=0$, such that $t=0$ or $t=1$. So which of these is actually the maximum? We can use the second derivative test (read more about that here). We have that the acceleration is $a(t)=6-12t$. Plugging in $t=0$, gives $a(t)=6$ which is positive and therefore a minimum. However, plugging in $t=1$ gives $-6$, which is negative and therefore, defines $t=1$ as a maximum.