"Kinematics" — Practice

Question 1

A particle moves 8 m east then 3 m west. Give the distance travelled and the magnitude of the displacement.

Accepted Answer

**Distance = 11 m** — the total path length, which can never decrease.

**Displacement = 5 m east** — the net change in position.

Check: distance $\geq$ |displacement|, as it must be whenever the motion reverses.

Question 2

State the slope and area interpretations of a velocity–time graph, and how each distinguishes displacement from distance.

Accepted Answer

**Slope** $= a$, the acceleration.

**Signed area** $= s$, the displacement — area below the axis counts negative.

**Total unsigned area** $=$ the distance travelled.

That sign is the whole difference between the two, which is why a graph question is usually faster than the algebra.

Question 3

A ball is thrown up at $30\ 	ext{m/s}$, with $g=10\ 	ext{m/s}^2$. Find the maximum height and the total time of flight.

Accepted Answer

$H=\dfrac{u^{2}}{2g}=\dfrac{900}{20}=\mathbf{45\ 	ext{m}}$

$T=\dfrac{2u}{g}=\dfrac{60}{10}=\mathbf{6\ 	ext{s}}$

Half of that, 3 s, is the rise time $u/g$ — and the ball returns to the launch level at $30\ 	ext{m/s}$ downward.

Question 4

A cyclist covers the first half of a distance at $30\ 	ext{km/h}$ and the second half at $20\ 	ext{km/h}$. Find the average speed.

Accepted Answer

Equal **distances** $\Rightarrow$ harmonic mean:

$v_{avg}=\dfrac{2v_1v_2}{v_1+v_2}=\dfrac{2(30)(20)}{50}=\dfrac{1200}{50}=\mathbf{24\ 	ext{km/h}}$

Below the arithmetic mean of 25, because more time is spent on the slower half. The 25 will be among the options.

Question 5

A body starts from rest with $a=4\ 	ext{m/s}^2$. Find its displacement in the 3rd second and verify by subtraction.

Accepted Answer

$s_3=u+\dfrac{a}{2}(2n-1)=0+2(5)=\mathbf{10\ 	ext{m}}$

*Verify:* $s_{3	ext{s}}=	frac12(4)(9)=18$ m and $s_{2	ext{s}}=	frac12(4)(4)=8$ m, difference $=10$ m.

The successive seconds give 2, 6, 10, 14 m — an AP with common difference $a=4$.

Question 6

A stone is thrown up at $15\ 	ext{m/s}$ from a $20\ 	ext{m}$ tower, $g=10\ 	ext{m/s}^2$. Find the time to reach the ground.

Accepted Answer

Up positive, origin at the throw $\Rightarrow$ ground at $s=-20$, $u=+15$, $a=-10$.

$-20=15t-5t^{2}\ \Rightarrow\ t^{2}-3t-4=0\ \Rightarrow\ (t-4)(t+1)=0$

$t=\mathbf{4\ 	ext{s}}$, rejecting $t=-1$ as pre-launch.

The stone is back at launch height at $t=3$ s, so the final 20 m takes only 1 s.

Question 7

A river flows at $4\ 	ext{m/s}$ and a boat moves at $5\ 	ext{m/s}$ in still water across a $90\ 	ext{m}$ width. Find the shortest crossing time and the resulting drift.

Accepted Answer

Aim straight across $\Rightarrow$ the full $5\ 	ext{m/s}$ is perpendicular.

$t=\dfrac{90}{5}=\mathbf{18\ 	ext{s}}$

Drift $=4	imes18=\mathbf{72\ 	ext{m}}$ downstream.

The current cannot affect the crossing time at all — it has no component across the river.

Question 8

For the same boat and river, find the steering angle needed to land directly opposite, and the time taken.

Accepted Answer

$5\sin	heta=4\ \Rightarrow\ \sin	heta=0.8\ \Rightarrow\ 	heta\approx\mathbf{53°}$ from the perpendicular, aimed upstream.

Effective crossing speed $=5\cos	heta=3\ 	ext{m/s}$, so $t=\dfrac{90}{3}=\mathbf{30\ 	ext{s}}$.

Longer than the 18 s above — the shortest path is never the shortest time.

Question 9

A projectile launched at $20\ 	ext{m/s}$ has a range of $20\ 	ext{m}$, with $g=10\ 	ext{m/s}^2$. Find both possible angles and compare the maximum heights.

Accepted Answer

$R=\dfrac{u^{2}\sin 2	heta}{g}\ \Rightarrow\ 20=\dfrac{400\sin 2	heta}{10}\ \Rightarrow\ \sin 2	heta=0.5$

$2	heta=30°$ or $150°$, so $	heta=\mathbf{15°}$ or $\mathbf{75°}$ — complementary, as always.

$H=\dfrac{u^{2}\sin^{2}	heta}{2g}$ gives $\mathbf{1.34\ 	ext{m}}$ and $\mathbf{18.7\ 	ext{m}}$.

Same range, nothing else in common: one flat and fast, one high and slow.

Question 10

Car A at $20\ 	ext{m/s}$ is $100\ 	ext{m}$ behind car B moving at $10\ 	ext{m/s}$ in the same direction. Using the relative frame, find when A catches B.

Accepted Answer

In B's frame, $v_{AB}=20-10=10\ 	ext{m/s}$ and A must close $100\ 	ext{m}$.

$t=\dfrac{100}{10}=\mathbf{10\ 	ext{s}}$

*Check in the ground frame:* A covers $200\ 	ext{m}$; B covers $100\ 	ext{m}$ from a start $100\ 	ext{m}$ ahead. Both are at $200\ 	ext{m}$ from A's start.

The relative frame turned a two-body problem into a one-body one — that is what it is for.

Question 11

Two vectors of equal magnitude $A$ have a resultant of magnitude $A$. Find the angle between them.

Accepted Answer

$A^{2}=A^{2}+A^{2}+2A^{2}\cos	heta$

$-A^{2}=2A^{2}\cos	heta \Rightarrow \cos	heta=-	frac12 \Rightarrow 	heta=\mathbf{120°}$

*Sanity check:* the resultant $A$ must lie between $|A-A|=0$ and $A+A=2A$, and it does. A resultant *smaller* than either vector requires an obtuse angle, so $120°$ is the right region.

This is the standard configuration of three equal coplanar vectors in equilibrium — each pair at $120°$, and the three summing to zero.

Question 12

A stone is projected horizontally with speed $u$ from a height. Find the radius of curvature of its path at the instant of projection.

Accepted Answer

At the moment of projection the velocity is horizontal, so gravity is entirely **perpendicular** to it. The whole of $g$ therefore acts as the centripetal acceleration:

$g=\dfrac{v^{2}}{R}$ with $v=u$

$$R=\frac{u^{2}}{g}$$

*Read it back:* a faster projectile has a **flatter** path, and $R	o\infty$ as $u	o\infty$ — a straight line, which is right.

*Later in the flight the answer changes.* Once the velocity has tilted, only the component of $g$ across the velocity bends the path, and the radius becomes $R=v^{3}/(u g)$ in general. At the topmost point of a full projectile that reduces to $u_x^{2}/g$.

Practice — "Kinematics"

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