Kinematics of Linear Motion, Worked Examples (KBAT)

First build the velocity by integrating the acceleration, using the initial velocity to fix the constant.

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

Since $v=5$ when $t=0$, we get $c=5$, so $v=t^{2}-6t+5$. Find where the particle is at rest.

$$t^{2}-6t+5=0\ \Rightarrow\ (t-1)(t-5)=0$$

So $v=0$ at $t=1$ and $t=5$. Checking signs, $v>0$ on $0\le t<1$ and $v<0$ on $1<t<5$; the particle reverses at $t=1$.

Integrate again for the displacement, with $s=0$ at $t=0$.

$$s=\int (t^{2}-6t+5)\,dt=\frac{t^{3}}{3}-3t^{2}+5t$$

Evaluate the displacement at the key times $t=1$ and $t=5$.

$$s(1)=\tfrac{1}{3}-3+5=\tfrac{7}{3},\qquad s(5)=\tfrac{125}{3}-75+25=-\tfrac{25}{3}$$

Add the distance of each leg as the absolute change in displacement.

$$\text{Total}=\left|\tfrac{7}{3}-0\right|+\left|-\tfrac{25}{3}-\tfrac{7}{3}\right|=\tfrac{7}{3}+\tfrac{32}{3}=\tfrac{39}{3}=13$$

Find the displacement of each particle from $O$ by integrating, using $s=0$ at $t=0$ for both.

$$s_{A}=\int (3t^{2}-6t)\,dt=t^{3}-3t^{2},\qquad s_{B}=\int 4\,dt=4t$$

(a) They meet where their displacements are equal.

$$t^{3}-3t^{2}=4t\ \Rightarrow\ t^{3}-3t^{2}-4t=0$$

Factorise, taking out $t$.

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

The relevant root is $t=4$, rejecting $t=0$ (the start) and $t=-1$. Their common displacement is:

$$s_{A}(4)=4^{3}-3(4)^{2}=64-48=16$$

(b) To get $A$’s total distance, find where $A$ is at rest, since $v_{A}=3t(t-2)$ changes sign.

$$3t(t-2)=0\ \Rightarrow\ t=0\ \text{or}\ t=2$$

$v_{A}<0$ on $0<t<2$ and $v_{A}>0$ on $2<t<4$, so $A$ reverses at $t=2$. Evaluate $s_{A}$.

$$s_{A}(2)=8-12=-4,\qquad s_{A}(4)=16$$

Add the two legs.

$$\text{Total}=|-4-0|+|16-(-4)|=4+20=24$$

Integrate the acceleration for the velocity, fixing the constant from $v=2$ at $t=0$.

$$v=\int (6-2t)\,dt=6t-t^{2}+c,\qquad v(0)=2\ \Rightarrow\ c=2$$

So $v=-t^{2}+6t+2$. The velocity is greatest where its rate of change, the acceleration, is zero.

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

Because $a>0$ before $t=3$ and $a<0$ after it, this is a maximum. Substitute $t=3$ into $v$.

$$v|_{t=3}=-(3)^{2}+6(3)+2=-9+18+2=11$$

(b) Integrate the velocity for the displacement, with $s=0$ at $t=0$.

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

Substitute $t=3$.

$$s|_{t=3}=-\frac{27}{3}+3(3)^{2}+2(3)=-9+27+6=24$$

Differentiate the displacement to get the velocity, since the particle is given here as a function of $s$ rather than of $a$.

$$v=\frac{ds}{dt}=3t^{2}-12t+9$$

(a) The particle is instantaneously at rest where $v=0$.

$$3t^{2}-12t+9=0\ \Rightarrow\ t^{2}-4t+3=0\ \Rightarrow\ (t-1)(t-3)=0$$

So the particle is at rest at $t=1$ and $t=3$. (b) To decide which instant gives the maximum displacement, differentiate again to get the acceleration and check its sign at each instant.

$$a=\frac{dv}{dt}=6t-12$$

$$a|_{t=1}=6(1)-12=-6<0,\qquad a|_{t=3}=6(3)-12=6>0$$

A negative acceleration at $t=1$ means $s$ is at a maximum there (turning from increasing to decreasing); a positive acceleration at $t=3$ means $s$ is at a minimum. Evaluate $s$ at $t=1$.

$$s|_{t=1}=(1)^{3}-6(1)^{2}+9(1)=1-6+9=4$$

Integrate the acceleration to get the velocity, using the given initial velocity to fix the constant.

$$v=\int (4t-8)\,dt=2t^{2}-8t+c,\qquad v(0)=6\ \Rightarrow\ c=6$$

So $v=2t^{2}-8t+6=2(t-1)(t-3)$. The particle is at rest at $t=1$ and $t=3$, both inside the first $4$ seconds, so it reverses direction twice.

Integrate again for the displacement, with $s=0$ at $t=0$.

$$s=\int (2t^{2}-8t+6)\,dt=\frac{2t^{3}}{3}-4t^{2}+6t$$

Evaluate the displacement at the start, at each rest instant, and at the end of the interval.

$$s(0)=0,\quad s(1)=\tfrac{8}{3},\quad s(3)=0,\quad s(4)=\tfrac{8}{3}$$

(a) Add the distance of each leg.

$$\text{Total distance}=\left|\tfrac{8}{3}-0\right|+\left|0-\tfrac{8}{3}\right|+\left|\tfrac{8}{3}-0\right|=\tfrac{8}{3}+\tfrac{8}{3}+\tfrac{8}{3}=8$$

(b) The average velocity uses only the net displacement over the time taken, not the distance covered.

$$\text{Average velocity}=\frac{s(4)-s(0)}{4}=\frac{\tfrac{8}{3}}{4}=\tfrac{2}{3}\ \text{m s}^{-1}$$

Find $A$'s velocity and displacement from $O$ by integrating twice, using $v_{A}=0$ and $s_{A}=0$ at $t=0$.

$$v_{A}=\int 6t\,dt=3t^{2},\qquad s_{A}=\int 3t^{2}\,dt=t^{3}$$

Since $B$ starts $16$ m from $O$ and moves towards $O$ at a constant $4\ \text{m s}^{-1}$, its displacement from $O$ decreases steadily.

$$s_{B}=16-4t$$

(a) $A$ and $B$ meet where their displacements from $O$ are equal.

$$t^{3}=16-4t\ \Rightarrow\ t^{3}+4t-16=0$$

Test small integer values: $t=2$ gives $8+8-16=0$, so $t=2$ is a root. Factorise it out.

$$t^{3}+4t-16=(t-2)(t^{2}+2t+8)=0$$

The quadratic factor has discriminant $2^{2}-4(1)(8)=-28<0$, so it has no real roots; $t=2$ is the only solution.

$$s_{A}(2)=2^{3}=8\ \text{m}$$

(b) Find each particle's velocity at $t=2$.

$$v_{A}(2)=3(2)^{2}=12\ \text{m s}^{-1},\qquad v_{B}=-4\ \text{m s}^{-1}\ (\text{constant})$$

(a) The particle is at rest where $v=0$.

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

(b) The minimum velocity occurs where the acceleration $a=\frac{dv}{dt}$ is zero, since $v$ is an upward-opening quadratic in $t$.

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

$$v|_{t=2}=(2)^{2}-4(2)+3=4-8+3=-1$$

So the minimum velocity is $-1\ \text{m s}^{-1}$ at $t=2$. (c) Speed is $|v|$, not $v$ itself, so check its value at the rest instants found in (a) as well.

$$|v|_{t=1}=0,\qquad |v|_{t=2}=1,\qquad |v|_{t=3}=0$$

Integrate the acceleration to get the velocity, using the given initial velocity to fix the constant of integration.

$$v=\int (4t-k)\,dt=2t^{2}-kt+c,\qquad v(0)=8\ \Rightarrow\ c=8$$

(a) Use the fact that $v=0$ at $t=1$ to find $k$.

$$v(1)=2(1)^{2}-k(1)+8=10-k=0\ \Rightarrow\ k=10$$

So $v=2t^{2}-10t+8$. Factorise to find the other instant at which the particle is at rest.

$$v=2(t^{2}-5t+4)=2(t-1)(t-4)\ \Rightarrow\ v=0\ \text{at}\ t=1\ \text{or}\ t=4$$

(b) Since the coefficient of $t^{2}$ is positive, $v$ is negative between its two roots. Test a value in between, say $t=2$, to confirm.

$$v(2)=2(2-1)(2-4)=2(1)(-2)=-4<0$$

Outside $[1,4]$ the factors $(t-1)$ and $(t-4)$ have the same sign, so $v>0$ there, matching the given $v(0)=8>0$.