SL 5.9—Kinematics problems

A particle is at rest when its velocity equals zero. We need to differentiate $s(t)$ to find $v(t)$, then solve $v(t)=0$.

$v(t)=\frac{ds}{dt}=6t-3t^2$

$6t-3t^2=0⟹3t(2-t)=0⟹t=0\text{ or }t=2$

Since the question asks for the time after $t=0$, the answer is $t=2$ s.

We need $v(t)=0$ where $v(t)=6t-3t^2=3t(2-t)$. Now test each option.

$v(1)=6(1)-3(1{)}^2=6-3=3\ne 0$. The particle is not at rest at $t=1$.

$v(2)=6(2)-3(4)=12-12=0$. The particle is at rest at $t=2$.

$v(3)=18-27=-9\ne 0$ and $v(6)=36-108\ne 0$. Neither gives $v=0$.

Only $t=2$ satisfies $v(t)=0$ for $t>0$, confirming the answer is $\mathbf{t}=2$ s.

We integrate $a(t)$ twice, applying initial conditions $v(0)=0$ (starts from rest) and $s(0)=2$ at each stage.

$v(t)=\int (6t-4)dt=3t^2-4t+C_1$ Using $v(0)=0$: $0=0-0+C_1\Rightarrow C_1=0$, so $v(t)=3t^2-4t$.

$s(t)=\int (3t^2-4t)dt=t^3-2t^2+C_2$ Using $s(0)=2$: $2=0-0+C_2\Rightarrow C_2=2$.

$s(t)=t^3-2t^2+2$

We need $s(0)=2$ and the function must result from integrating $a(t)=6t-4$ twice with $v(0)=0$.

$s(0)=0\ne 2$. This fails the initial condition $s(0)=2$.

This would give $v(t)=6t-4=a(t)$, meaning this is the result of only one integration — not two. It does not satisfy $v(0)=0$: $v(0)=-4\ne 0$.

Differentiating gives $v(t)=6t^2-8t$, and then $a(t)=12t-8\ne 6t-4$. Incorrect.

$s(t)=t^3-2t^2+2$ satisfies $s(0)=2$, and its derivatives give $v(t)=3t^2-4t$ (so $v(0)=0$ ✓) and $a(t)=6t-4$ ✓.