Kinematics — Practice Questions with Answers

Consider a $v$–$t$ graph where velocity increases uniformly from $u$ (at $t=0$) to $v$ (at time $t$), forming a straight line.

The slope of this line represents acceleration: $a = \dfrac{v - u}{t - 0}$

Rearranging: $at = v - u$

$\therefore v = u + at$

Scalar quantities are completely described by magnitude alone, with no associated direction. Examples: mass, speed, distance, time, temperature.

Vector quantities require both magnitude and direction to be completely specified, and follow the laws of vector addition. Examples: displacement, velocity, acceleration, force.

A key distinguishing test: vector quantities can have components resolved along different directions and combine using vector addition (not simple arithmetic), while scalars combine using ordinary algebraic addition.

Horizontal range $R$ = horizontal velocity × time of flight.

Horizontal velocity = $u\cos\theta$ (constant, since there's no horizontal acceleration)

Time of flight $T = \dfrac{2u\sin\theta}{g}$

$R = u\cos\theta \times \dfrac{2u\sin\theta}{g} = \dfrac{2u^2\sin\theta\cos\theta}{g} = \dfrac{u^2\sin2\theta}{g}$

(using the identity $2\sin\theta\cos\theta = \sin2\theta$)

Relative velocity is the velocity of one object as observed from the reference frame of another moving object — mathematically, $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$.

Example: If two trains move in the same direction at 60 km/h and 40 km/h, a passenger on the faster train sees the slower train moving backward at $60-40=20$ km/h. If they move in opposite directions at these same speeds, the relative velocity becomes $60+(-(-40))=100$ km/h — they approach each other much faster.

Vertical component of initial velocity = $u\sin\theta$ (upward), with downward acceleration $g$.

At the point where the projectile returns to the same vertical level, net vertical displacement = 0. Using $s = ut + \frac{1}{2}at^2$ for the vertical direction:

$0 = (u\sin\theta)T - \frac{1}{2}gT^2$

$T\left[u\sin\theta - \frac{1}{2}gT\right] = 0 \Rightarrow T = \dfrac{2u\sin\theta}{g}$ (rejecting the trivial solution $T=0$)

Average velocity = total displacement / total time (a vector quantity).

Average speed = total distance travelled / total time (a scalar quantity).

Yes — average speed can be greater than the magnitude of average velocity whenever the path is not a straight line, since distance travelled is always greater than or equal to the magnitude of displacement. For example, if a body returns to its starting point, average velocity is zero, but average speed is clearly non-zero.

At maximum height, the vertical component of velocity becomes zero. Using $v^2 = u^2 - 2as$ for the vertical direction, with initial vertical velocity $u\sin\theta$:

$0 = (u\sin\theta)^2 - 2gH_{max}$

$H_{max} = \dfrac{u^2\sin^2\theta}{2g}$

For a body starting with initial velocity $u$, moving with uniform acceleration $a$, reaching velocity $v$ after time $t$, and covering displacement $s$:

1. First equation: $v = u + at$
2. Second equation: $s = ut + \dfrac{1}{2}at^2$
3. Third equation: $v^2 = u^2 + 2as$

Uniform circular motion is motion along a circular path at constant speed (the magnitude of velocity does not change).

However, velocity is a vector — even though its magnitude (speed) stays constant, its direction continuously changes as the object moves around the circle. Since acceleration is defined as the rate of change of velocity (a vector), this continuous change in direction means the object is always accelerating, even though its speed never changes. This acceleration — called centripetal acceleration — always points towards the centre of the circle.

Acceleration $a = \dfrac{dv}{dt}$, so $dv = a\,dt$. Integrating from $0$ to $t$ (velocity goes from $u$ to $v$):

$v = u + at$ ... (i)

Also, $v = \dfrac{ds}{dt}$, so $ds = v\,dt = (u+at)\,dt$ (using equation i).

Integrating both sides from $t=0$ to $t$ (displacement goes from $0$ to $s$):

$s = ut + \dfrac{1}{2}at^2$

Given: $u = 20\text{ m/s}$, $\theta = 30°$, $g = 10\text{ m/s}^2$. Note $\sin30° = 0.5$, $\cos30° = 0.866$.

(a) Time of flight: $T = \dfrac{2u\sin\theta}{g} = \dfrac{2\times20\times0.5}{10} = \dfrac{20}{10} = 2\text{ s}$

(b) Maximum height: $H = \dfrac{u^2\sin^2\theta}{2g} = \dfrac{400\times0.25}{20} = \dfrac{100}{20} = 5\text{ m}$

(c) Horizontal range: $R = \dfrac{u^2\sin2\theta}{g} = \dfrac{400\times\sin60°}{10} = \dfrac{400\times0.866}{10} \approx 34.6\text{ m}$

Given: $u = 0$, $a = 2\text{ m/s}^2$, $t = 10\text{ s}$.

(a) Final velocity: Using $v = u + at$: $v = 0 + (2)(10) = 20\text{ m/s}$

(b) Distance covered: Using $s = ut + \frac{1}{2}at^2$: $s = 0 + \frac{1}{2}(2)(10)^2 = \frac{1}{2}(2)(100) = 100\text{ m}$

Verification using third equation: $v^2 = u^2 + 2as \Rightarrow (20)^2 = 0 + 2(2)(100) = 400$ ✓ Confirmed.

Relative velocity: Since both move in the same direction, $v_{AB} = v_A - v_B = 90 - 60 = 30\text{ km/h}$

Converting to m/s: $30\text{ km/h} = 30 \times \dfrac{5}{18} = \dfrac{25}{3}\text{ m/s} \approx 8.33\text{ m/s}$

Time to catch up: In B's reference frame, A must cover the 500 m gap at the relative velocity of 8.33 m/s.

$t = \dfrac{\text{distance}}{\text{relative velocity}} = \dfrac{500}{8.33} \approx 60\text{ s}$

Car A takes approximately 60 seconds (1 minute) to catch up with car B.

Given: $u = 0$ (dropped, not thrown), $h = 45\text{ m}$, $g = 10\text{ m/s}^2$.

(a) Time taken: Using $h = ut + \frac{1}{2}gt^2$: $45 = 0 + \frac{1}{2}(10)t^2 = 5t^2$

$t^2 = 9 \Rightarrow t = 3\text{ s}$

(b) Final velocity: Using $v = u + gt$: $v = 0 + (10)(3) = 30\text{ m/s}$

The stone takes 3 seconds to fall and strikes the ground at 30 m/s.

Given: $u = 30\text{ m/s}$, $g = 10\text{ m/s}^2$ (deceleration while rising).

(a) Maximum height: At max height, $v = 0$. Using $v^2 = u^2 - 2gh$: $0 = 900 - 2(10)h \Rightarrow h = \dfrac{900}{20} = 45\text{ m}$

(b) Time to reach maximum height: Using $v = u - gt$: $0 = 30 - 10t \Rightarrow t = 3\text{ s}$

(c) Total time in air: By symmetry, time to fall back equals time to rise, so total time = $2 \times 3 = 6\text{ s}$

Given: $r = 0.5\text{ m}$, $v = 4\text{ m/s}$.

Centripetal acceleration: $a_c = \dfrac{v^2}{r} = \dfrac{(4)^2}{0.5} = \dfrac{16}{0.5} = 32\text{ m/s}^2$

Time period: Circumference = $2\pi r = 2\pi(0.5) = \pi\text{ m} \approx 3.14\text{ m}$

$T = \dfrac{\text{circumference}}{v} = \dfrac{3.14}{4} \approx 0.785\text{ s}$

Consider a projectile launched with speed $u$ at angle $\theta$. At time $t$:

Horizontal displacement: $x = (u\cos\theta)t \Rightarrow t = \dfrac{x}{u\cos\theta}$ ... (i)

Vertical displacement: $y = (u\sin\theta)t - \dfrac{1}{2}gt^2$ ... (ii)

Substituting (i) into (ii):

$y = (u\sin\theta)\dfrac{x}{u\cos\theta} - \dfrac{1}{2}g\left(\dfrac{x}{u\cos\theta}\right)^2$

$y = x\tan\theta - \dfrac{g}{2u^2\cos^2\theta}x^2$

This equation has the form $y = Ax - Bx^2$, where $A = \tan\theta$ and $B = \dfrac{g}{2u^2\cos^2\theta}$ are constants — this is precisely the equation of a parabola, confirming that projectile motion traces a parabolic path.

Converting speeds to m/s: $54\text{ km/h} = 54 \times \dfrac{5}{18} = 15\text{ m/s}$; $36\text{ km/h} = 36 \times \dfrac{5}{18} = 10\text{ m/s}$

Relative velocity (opposite directions, so speeds add): $v_{rel} = 15 + 10 = 25\text{ m/s}$

Total distance to be covered for complete crossing = sum of both train lengths = $300 + 200 = 500\text{ m}$

$t = \dfrac{\text{distance}}{\text{relative velocity}} = \dfrac{500}{25} = 20\text{ s}$

The trains take 20 seconds to completely cross each other.

The $v$–$t$ graph consists of three regions — a rising triangle, a rectangle, and a falling triangle. Total distance = total area under the graph.

Phase 1 (0 to 4 s, triangle): Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(4)(20) = 40\text{ m}$

Phase 2 (4 to 10 s, rectangle): Area = base × height = $(6)(20) = 120\text{ m}$

Phase 3 (10 to 15 s, triangle): Area = $\frac{1}{2}(5)(20) = 50\text{ m}$

Total distance = $40 + 120 + 50 = \mathbf{210\text{ m}}$

Let both balls meet at height $h$ from the ground, after time $t$, measured from the ground.

Ball 1 (dropped from top): Distance fallen from top = $\frac{1}{2}gt^2$, so its height above ground = $100 - \frac{1}{2}gt^2$

Ball 2 (thrown up from ground): Height above ground = $25t - \frac{1}{2}gt^2$

Setting the two heights equal (they meet at the same height at the same time):

$100 - \frac{1}{2}gt^2 = 25t - \frac{1}{2}gt^2$

The $\frac{1}{2}gt^2$ terms cancel: $100 = 25t \Rightarrow t = 4\text{ s}$

Height at meeting point: $h = 25(4) - \frac{1}{2}(10)(4)^2 = 100 - 80 = 20\text{ m}$

The two balls meet after 4 seconds, at a height of 20 m above the ground.