"Kinematics" — Practice

Question 1

A particle moves along a line with $v=4t^{3}-2t$ (SI units). Find its acceleration at $t=2\ 	ext{s}$ and its displacement in the first $2\ 	ext{s}$.

Accepted Answer

$a=\dfrac{dv}{dt}=12t^{2}-2$, so $a(2)=48-2=\mathbf{46\ 	ext{m/s}^{2}}$

$x=\displaystyle\int_0^2\left(4t^{3}-2tight)dt=\left[t^{4}-t^{2}ight]_0^2=16-4=\mathbf{12\ 	ext{m}}$

*Note why no standard formula applies:* the acceleration is $12t^{2}-2$, which changes from $-2$ to $46\ 	ext{m/s}^{2}$ over these two seconds. Averaging it and substituting would be meaningless.

Question 2

Two trains of lengths $100\ 	ext{m}$ and $150\ 	ext{m}$ travel in opposite directions on parallel tracks at $20\ 	ext{m/s}$ and $30\ 	ext{m/s}$. How long do they take to pass each other completely?

Accepted Answer

In the first train's frame the second approaches at

$v_{rel}=20+30=50\ 	ext{m/s}$

To pass completely, the relative displacement must be the **sum of the lengths**, since the front of one starts level with the front of the other and must end level with its rear:

$d=100+150=250\ 	ext{m}$

$t=\dfrac{250}{50}=\mathbf{5\ 	ext{s}}$

*The two standard errors:* subtracting the speeds (they are moving oppositely, so the relative speed is the sum), and using only one length.

Question 3

A boat crosses a river of width $d$ at speed $v$ in still water, with a current $u$. State the steering direction for the shortest crossing time, the time achieved, and the resulting drift.

Accepted Answer

The crossing time is fixed by the component **perpendicular to the bank**, which is $v\cos	heta$ for a steering angle $	heta$ from the normal. This is largest at $	heta=0$.

**Point the boat straight across.**

$t_{min}=\dfrac{d}{v}, \qquad 	ext{drift}=u\,t_{min}=\dfrac{ud}{v}$

*The trade-off:* this is the fastest crossing and also the one with the largest drift. Any upstream steering reduces the across-component and therefore lengthens the crossing — there is no angle that improves both.

Question 4

A particle starts from rest at the origin and moves along the $x$-axis with $a=2x+3$ (SI units). Find its speed at $x=2\ 	ext{m}$.

Accepted Answer

Use the chain-rule form $a=v\dfrac{dv}{dx}$:

$\displaystyle\int_0^{v}v\,dv=\int_0^{2}(2x+3)\,dx$

$\dfrac{v^{2}}{2}=\left[x^{2}+3xight]_0^{2}=4+6=10$

$v^{2}=20 \Rightarrow v=\mathbf{4.47\ 	ext{m/s}}$

*What this really was:* multiplying through by $m$ turns the left side into kinetic energy and the right into $\int F\,dx$. The chain-rule form of acceleration **is** the work-energy theorem, which is why it always works and why it never needs the time.

Question 5

At $t=0$, particle A is at the origin with velocity $(2,0)\ 	ext{m/s}$ and particle B is at $(10,12)\ 	ext{m}$ with velocity $(0,-2)\ 	ext{m/s}$. Find the least distance between them and the instant at which it occurs.

Accepted Answer

$\vec{r}_{AB}=(0,0)-(10,12)=(-10,-12)$

$\vec{v}_{AB}=(2,0)-(0,-2)=(2,2)$

$\vec{r}_{AB}\cdot\vec{v}_{AB}=-20-24=-44<0$, so they are still approaching.

$t_{min}=-\dfrac{-44}{|(2,2)|^{2}}=\dfrac{44}{8}=\mathbf{5.5\ 	ext{s}}$

$\left|\vec{r}_{AB}	imes\vec{v}_{AB}ight|=|(-10)(2)-(-12)(2)|=|-20+24|=4$

$d_{min}=\dfrac{4}{2\sqrt2}=\mathbf{1.41\ 	ext{m}}$

*Direct check:* at $t=5.5$ s, A is at $(11,0)$ and B at $(10,1)$. Separation $=\sqrt{1+1}=1.41$ m. Confirmed.

Question 6

To a man walking at $3\ 	ext{m/s}$ the rain appears to fall **vertically**. When he doubles his speed it appears to come at $45°$ to the vertical. Find the actual speed and direction of the rain.

Accepted Answer

Let $\vec{v}_r=(a,-w)$ with the man walking along $+x$.

**First observation.** $\vec{v}_{rm}=(a-3,-w)$ is vertical, so its horizontal part vanishes:

$a=3\ 	ext{m/s}$ — the rain is already drifting forward at the man's original speed.

**Second observation.** At $6\ 	ext{m/s}$, $\vec{v}_{rm}=(3-6,-w)=(-3,-w)$, and $45°$ to the vertical means the two components are equal in size:

$w=3\ 	ext{m/s}$

$$\vec{v}_r=(3,-3),\qquad |\vec{v}_r|=3\sqrt2=\mathbf{4.24\ 	ext{m/s}}$$

at $\mathbf{45°}$ to the vertical, blowing in the direction the man walks.

*The insight:* rain appearing vertical does **not** mean it is falling vertically. It means the observer happens to be matching its horizontal drift.

Question 7

A particle is projected at $20\ 	ext{m/s}$ at $45°$ to a $15°$ incline, up the slope. Take $g=10$. Find the time of flight and the range along the incline.

Accepted Answer

With $\alpha=45°$ measured from the incline and $\beta=15°$:

$T=\dfrac{2u\sin\alpha}{g\cos\beta}=\dfrac{2(20)(0.707)}{10(0.966)}=\mathbf{2.93\ 	ext{s}}$

$R=\dfrac{2u^{2}\sin\alpha\cos(\alpha+\beta)}{g\cos^{2}\beta}=\dfrac{2(400)(0.707)\cos 60°}{10(0.933)}$

$R=\dfrac{565.7(0.5)}{9.33}=\mathbf{30.3\ 	ext{m}}$

*Was $45°$ optimal?* The optimum is $\alpha=(90°-15°)/2=37.5°$, so no. At that angle $R_{max}=\dfrac{u^{2}}{g(1+\sin 15°)}=\dfrac{400}{12.59}=31.8$ m — only $1.5$ m better, because the range varies slowly near its maximum.

Question 8

A particle moves with position vector $\vec{r}=2t\,\hat{i}+t^{2}\,\hat{j}$ (SI units). Find the radius of curvature of its path at $t=1\ 	ext{s}$.

Accepted Answer

$\vec{v}=\dfrac{d\vec{r}}{dt}=2\hat{i}+2t\,\hat{j}$, so at $t=1$: $\vec{v}=(2,2)$ and $|\vec{v}|=2\sqrt2$

$\vec{a}=\dfrac{d\vec{v}}{dt}=2\hat{j}=(0,2)$, constant

$\left|\vec{v}	imes\vec{a}ight|=|2(2)-2(0)|=4$

$R=\dfrac{\left|\vec{v}ight|^{3}}{\left|\vec{v}	imes\vec{a}ight|}=\dfrac{(2\sqrt2)^{3}}{4}=\dfrac{16\sqrt2}{4}=4\sqrt2=\mathbf{5.66\ 	ext{m}}$

*Recognise the path:* $x=2t$ and $y=t^{2}$ give $y=x^{2}/4$ — a parabola, with constant acceleration in $+y$. It is a projectile problem written in vector notation, upside down.

Question 9

A gun fires a bullet at speed $u$ aimed **directly at** a target sitting at horizontal distance $R$ and height $H$. At the instant of firing, the target begins to fall freely. Find the minimum $u$ for which the bullet strikes the target before it reaches the ground.

Accepted Answer

Bullet and target share the acceleration $\vec{g}$, so $\vec{a}_{rel}=0$ and **in the target's frame the bullet moves in a straight line at constant speed $u$**, straight along the original line of sight. A hit is therefore guaranteed in principle — the only question is whether it happens in time.

**Time for the bullet to cover the line of sight:**

$L=\sqrt{R^{2}+H^{2}}, \qquad t_{hit}=\dfrac{L}{u}$

**Time for the target to fall $H$:**

$t_{fall}=\sqrt{\dfrac{2H}{g}}$

Requiring $t_{hit}\le t_{fall}$:

$$u\ \ge\ \sqrt{R^{2}+H^{2}}\,\sqrt{\frac{g}{2H}}$$

*Read the structure:* the aiming is automatically correct at **every** speed — that is what zero relative acceleration buys. The speed condition is not about accuracy at all, only about arriving before the target lands.

Question 10

A boat of mass $m$ moving at $v_0$ shuts off its engine. The water resists with $F=-bv^{2}$. Find $v$ as a function of distance, and compare the total distance travelled with the linear-resistance case $F=-kmv$.

Accepted Answer

$m\,v\dfrac{dv}{dx}=-bv^{2} \Rightarrow \dfrac{dv}{v}=-\dfrac{b}{m}\,dx$

$$v=v_0e^{-bx/m}$$

**The boat never stops, and it travels an infinite distance.** $v$ reaches zero only as $x	o\infty$.

**Contrast with linear resistance.** There $v\,dv/dx=-kv$ gives $dv/dx=-k$, so

$v=v_0-kx$, and $v=0$ at the **finite** distance $x=\dfrac{v_0}{k}$.

| Resistance | $v(x)$ | Stops at |
|---|---|---|
| $-kmv$ | $v_0-kx$ | $x=v_0/k$, finite |
| $-bv^{2}$ | $v_0e^{-bx/m}$ | never |

*Why the reversal:* quadratic drag is fierce at high speed but collapses at low speed, so the boat's last metres are almost unresisted and it creeps on forever. Linear drag stays proportionally strong all the way down and brings it to a halt.

Question 11

A particle is projected up an incline of angle $\beta$ at angle $\alpha$ to the incline surface. Find the condition for it to strike the incline **perpendicularly**.

Accepted Answer

Take axes along and perpendicular to the incline. Striking the surface at right angles means the **along-slope** velocity component has fallen to zero at landing.

Along the slope: $u_{\parallel}=u\cos\alpha$ with deceleration $g\sin\beta$.

Perpendicular to it: the flight time is set as usual by the component that starts and ends at zero,

$T=\dfrac{2u\sin\alpha}{g\cos\beta}$

Imposing $v_{\parallel}=0$ at $t=T$:

$u\cos\alpha-g\sin\beta\left(\dfrac{2u\sin\alpha}{g\cos\beta}ight)=0$

$\cos\alpha\cos\beta=2\sin\alpha\sin\beta$

$$\cot\alpha=2	an\beta$$

*Check the limits.* At $\beta=0$ (flat ground) this gives $\cot\alpha=0$, so $\alpha=90°$ — thrown straight up, landing straight down, perpendicular to the ground. Correct. As $\beta$ grows, $\alpha$ shrinks: on a steep slope you must throw much closer to the surface.

Question 12

A river of width $d$ flows at $u$, and a boat does $v$ in still water with $u>v$. Derive the minimum possible drift and the steering angle that achieves it.

Accepted Answer

Steering at $	heta$ upstream from the normal:

$	ext{across} = v\cos	heta \Rightarrow t=\dfrac{d}{v\cos	heta}$

$	ext{drift}=\left(u-v\sin	hetaight)t=\dfrac{d\left(u-v\sin	hetaight)}{v\cos	heta}$

Since $u>v$, the numerator can never reach zero — **no steering angle eliminates the drift**. Minimise instead:

$\dfrac{d}{d	heta}\left(\dfrac{u-v\sin	heta}{\cos	heta}ight)=\dfrac{-v\cos^{2}	heta+\left(u-v\sin	hetaight)\sin	heta}{\cos^{2}	heta}=0$

$-v\cos^{2}	heta+u\sin	heta-v\sin^{2}	heta=0 \Rightarrow u\sin	heta=v$

$$\sin	heta=\frac{v}{u}$$

Substituting back, with $\cos	heta=\dfrac{\sqrt{u^{2}-v^{2}}}{u}$:

$$	ext{drift}_{min}=\frac{d\sqrt{u^{2}-v^{2}}}{v}$$

*Note the pairing.* When the boat is faster, zero drift needs $\sin	heta=u/v$; when the river is faster, least drift needs $\sin	heta=v/u$. **Reciprocals** — and only one of them is a legal sine in any given problem, which is a useful check that you picked the right case.

Practice — "Kinematics"

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