Kinematics of Motion in a Straight Line: Question 9

Worked solution

Part (a): Finding velocity and acceleration

Since $v = \dfrac{ds}{dt}$, differentiate $s$ with respect to $t$:

$v = \frac{d}{dt}\left(t^3-6t^2+9t\right) = 3t^2 - 12t + 9$

Since $a = \dfrac{dv}{dt}$, differentiate $v$ with respect to $t$:

$a = \frac{d}{dt}\left(3t^2-12t+9\right) = 6t - 12$

Recompute independently as a check: applying the power rule term by term to $s$: the derivative of $t^3$ is $3t^2$, of $-6t^2$ is $-12t$, and of $9t$ is $9$, giving $v=3t^2-12t+9$ (matching. Differentiating $v$ term by term: the derivative of $3t^2$ is $6t$, of $-12t$ is $-12$, and of the constant $9$ is $0$, giving $a=6t-12$) matching.

Part (b): Finding when $P$ is instantaneously at rest

$P$ is at rest when $v=0$:

$3t^2 - 12t + 9 = 0$

Dividing throughout by $3$:

$t^2 - 4t + 3 = 0$

Factorising:

$(t-1)(t-3) = 0$

So $t=1$ or $t=3$.

Recompute independently as a check: expanding $(t-1)(t-3) = t^2-3t-t+3 = t^2-4t+3$, which matches the equation above. Substituting back into $v=3t^2-12t+9$: at $t=1$, $v=3-12+9=0$ ✓; at $t=3$, $v=27-36+9=0$ ✓. Both confirm $P$ is at rest at $t=1$ and $t=3$.

Part (c): Acceleration at each rest point, and direction changes

Using $a=6t-12$:

At $t=1$: $a = 6(1)-12 = -6$

At $t=3$: $a = 6(3)-12 = 6$

At a rest point, if the acceleration is non-zero, the velocity is genuinely changing sign there (not just touching zero), so $P$ does change direction.

• At $t=1$: $a=-6\ne0$, so $P$ changes direction. Just before $t=1$, $v>0$ (e.g. $v(0)=9>0$), and just after, $v<0$ (e.g. $v(2)=3(4)-24+9=-3<0$), confirming $P$ switches from moving in the positive direction to the negative direction.
• At $t=3$: $a=6\ne0$, so $P$ changes direction again. Just before $t=3$, $v<0$ (as at $t=2$ above), and just after, $v>0$ (e.g. $v(4)=3(16)-48+9=9>0$), confirming $P$ switches back to moving in the positive direction.

Final answers

• (a) $v = \boxed{3t^2-12t+9}$, $a = \boxed{6t-12}$
• (b) $P$ is instantaneously at rest at $t=\boxed{1}$ and $t=\boxed{3}$
• (c) At $t=1$, $a=\boxed{-6\text{ m s}^{-2}}$ ($P$ changes direction); at $t=3$, $a=\boxed{6\text{ m s}^{-2}}$ ($P$ changes direction)