Kinematics of Linear Motion, Worked Examples (medium)

Total distance differs from displacement whenever the particle changes direction, so first find where the velocity is zero.

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

Divide by $3$ and factorise.

$$t^{2}-4t+3=0\ \Rightarrow\ (t-1)(t-3)=0$$

So $v=0$ at $t=1$ and $t=3$. Testing a value in each interval, $v>0$ on $0\le t<1$ and $v<0$ on $1<t<3$, so the particle reverses direction at $t=1$.

Find the displacement function by integrating, taking $s=0$ at $t=0$.

$$s=\int (3t^{2}-12t+9)\,dt=t^{3}-6t^{2}+9t$$

Evaluate the displacement at $t=0$, $t=1$ and $t=3$.

$$s(0)=0,\quad s(1)=1-6+9=4,\quad s(3)=27-54+27=0$$

The particle moves from $0$ to $4$ m in the first second, then back from $4$ m to $0$. Add the distance of each leg.

$$\text{Total distance}=|4-0|+|0-4|=4+4=8$$

Differentiate the displacement to get the velocity.

$$v=\frac{ds}{dt}=3t^{2}-18t+24$$

(a) The particle is at rest when $v=0$. Divide by $3$ and factorise.

$$t^{2}-6t+8=0\ \Rightarrow\ (t-2)(t-4)=0$$

So the particle is momentarily at rest at $t=2$ s and $t=4$ s.

(b) The velocity is least where its rate of change is zero, so set the acceleration to zero.

$$a=\frac{dv}{dt}=6t-18=0\ \Rightarrow\ t=3$$

Since $a<0$ before $t=3$ and $a>0$ after it, the velocity has a minimum there. Substitute $t=3$ into $v$.

$$v|_{t=3}=3(3)^{2}-18(3)+24=27-54+24=-3$$

Integrate the acceleration to get the velocity, keeping a constant $c_{1}$.

$$v=\int (6t-2)\,dt=3t^{2}-2t+c_{1}$$

Use $v=3$ when $t=1$.

$$3=3(1)^{2}-2(1)+c_{1}=1+c_{1}\ \Rightarrow\ c_{1}=2$$

So $v=3t^{2}-2t+2$. The initial velocity is its value at $t=0$.

$$v|_{t=0}=3(0)^{2}-2(0)+2=2$$

(b) Integrate the velocity to get the displacement, keeping a constant $c_{2}$.

$$s=\int (3t^{2}-2t+2)\,dt=t^{3}-t^{2}+2t+c_{2}$$

Use $s=5$ when $t=1$.

$$5=1-1+2+c_{2}=2+c_{2}\ \Rightarrow\ c_{2}=3$$

So $s=t^{3}-t^{2}+2t+3$. Substitute $t=2$.

$$s|_{t=2}=(2)^{3}-(2)^{2}+2(2)+3=8-4+4+3=11$$

The particle is furthest from $O$ at the instant it turns back, so first find where the velocity is zero.

$$8-4t=0\ \Rightarrow\ t=2$$

Integrate the velocity to get the displacement function, taking $s=0$ at $t=0$.

$$s=\int (8-4t)\,dt=8t-2t^{2}$$

Since $v>0$ for $0\le t<2$ and $v<0$ for $t>2$, the particle moves away from $O$ then returns, so $t=2$ gives the furthest point. Substitute $t=2$.

$$s|_{t=2}=8(2)-2(2)^{2}=16-8=8$$

The particle passes through $O$ whenever $s=0$, so factorise the displacement.

$$s=t(t^{2}-6t+8)=t(t-2)(t-4)$$

So $s=0$ when $t=0$, $t=2$ or $t=4$. The particle passes through $O$ again at $t=2$ s and $t=4$ s.

(b) Differentiate to get the velocity, then evaluate it at the later time, $t=4$.

$$v=\frac{ds}{dt}=3t^{2}-12t+8$$

$$v|_{t=4}=3(4)^{2}-12(4)+8=48-48+8=8$$

First find the displacement function by integrating the velocity, taking $s=0$ at $t=0$.

$$s=\int (2t-6)\,dt=t^{2}-6t$$

Check whether the particle changes direction inside $[1,4]$ by solving $v=0$.

$$2t-6=0\ \Rightarrow\ t=3$$

Since $t=3$ lies inside the interval, split it there and evaluate $s$ at $t=1$, $t=3$ and $t=4$.

$$s(1)=1-6=-5,\quad s(3)=9-18=-9,\quad s(4)=16-24=-8$$

$$\text{Total distance}=|{-9}-({-5})|+|{-8}-({-9})|=4+1=5$$

Speed is the magnitude of velocity, so the speed is $5$ when $v=5$ or $v=-5$. Try $v=5$ first.

$$t^{2}-2t-3=5\ \Rightarrow\ t^{2}-2t-8=0\ \Rightarrow\ (t-4)(t+2)=0$$

This gives $t=4$ or $t=-2$; reject $t=-2$ since $t\ge 0$. Now try $v=-5$.

$$t^{2}-2t-3=-5\ \Rightarrow\ t^{2}-2t+2=0$$

The discriminant of this equation is $(-2)^{2}-4(1)(2)=-4$, which is negative, so there is no real value of $t$ for this case.

The particle speeds up when its velocity and acceleration have the same sign, and slows down when they have opposite signs. Differentiate to get the velocity.

$$v=\frac{ds}{dt}=3t^{2}-8t$$

Differentiate again to get the acceleration, then evaluate both at $t=1$.

$$a=\frac{dv}{dt}=6t-8$$

$$v|_{t=1}=3(1)^{2}-8(1)=-5,\qquad a|_{t=1}=6(1)-8=-2$$