Displacement measures the straight-line change in position and includes direction, so it is a vector. Speed, distance and time are scalars (no direction).

Net displacement = final position − initial position = (3 m east − 4 m west) = 1 m west (because the 4 m west dominates by 1 m).

Velocity includes both magnitude (speed) and the direction of motion — therefore it is a vector. Speed and distance are scalars; time is scalar.

Instantaneous velocity is the instantaneous rate of change of displacement — graphically the slope of the tangent to s–t curve. Area under s–t doesn't have common kinematic meaning; area under v–t gives displacement.

Acceleration means change in velocity (magnitude or direction). Changes in speed or direction are acceleration. A color change has nothing to do with motion, so it does not imply acceleration.

Distance is the total path length and is always greater than or equal to the magnitude of displacement (which is shortest straight-line separation). They are equal only for straight-line one-direction motion.

Acceleration is change in velocity per unit time; SI unit is metres per second squared (m/s²).

Convert by dividing by 3.6: 72 ÷ 3.6 = 20 m/s (or multiply by 5/18).

A straight s–t line with constant slope indicates constant velocity i.e., uniform motion. Rest would be horizontal line (s constant).

Constant velocity means no change in velocity, so acceleration (rate of change of velocity) is zero.

Displacement has magnitude and direction (vector). Distance and speed are scalars; time is scalar.

m (metre) is a unit of distance, not speed. Speed requires distance per unit time units like m/s, km/h, cm/s.

This is the fundamental equation linking initial velocity u, final velocity v, acceleration a and time t for constant a. Options A and C have signs reversed and D is similar but sign depends on convention; standard forms list v² = u² + 2as.

If u = 0 (starts from rest), s = ut + ½at² reduces to s = ½at².

v² = u² + 2as links velocities, acceleration and displacement without time — useful when time is not provided.

a = (v − u)/t = (30 − 10)/5 = 20/5 = 4 m/s².

s = ut + ½at² = 5×4 + 0.5×2×16 = 20 + 16 = 36 — wait, that's 36. But 0.5×2×16 = 16, so 20+16=36. Option D is 36 m. (Correct option should be D.)

Equal distances in equal times means constant speed/velocity (no change), so acceleration is zero — this is uniform velocity, not uniform acceleration.

For deceleration, v=0 → 0 = u² + 2as ⇒ s = −u²/(2a). Since a is negative for deceleration, s = u²/(2|a|). Hence u²/2|a| (positive distance).

s = ut + ½at² includes initial velocity u, acceleration a and time t but not final velocity v explicitly (though v is related via v = u + at).

From v = u + at with u = 0, v = at.

This equation relates u, v, a, s and does not include time, so it’s used when time is not available.

Area under velocity–time curve (velocity × time) equals displacement over the time interval.

Horizontal s–t line means displacement does not change with time → object is at rest (velocity = 0).

Slope of velocity–time graph (Δv/Δt) gives acceleration.

When v–t line crosses the time axis, velocity equals zero at that instant; negative slope indicates uniform negative acceleration (deceleration).

Area under velocity-time curve equals displacement over that time interval (taking direction sign into account).

s = ut + ½at² yields a parabolic s–t curve when acceleration is constant. Upward opening indicates increasing slope (increasing velocity) — uniform acceleration.

Speed is scalar and non-negative; instantaneous speed cannot be negative. Velocity can be negative (directional).

If the v–t curve remains above time-axis (positive velocity) but slope is negative, velocity is decreasing — deceleration but still positive velocity values.

Near Earth’s surface, g ≈ 9.8 m/s². For school problems g ≈ 9.8 or sometimes 10 m/s² for approximation.

Gravity acts downward always; acceleration due to gravity (−g if upward positive) remains constant during ascent, at top and during descent (ignoring air resistance). Velocity becomes zero at top, but acceleration is still −g.

From v = u + at with u = 0 and a = g, v = gt (downwards magnitude gt).

Acceleration due to gravity is independent of initial speed or mass (neglecting air resistance). Both experience acceleration g downward.

For symmetric motion under constant gravity and equal launch/landing heights, ascent time equals descent time. Total flight time = 2 × ascent time.

Using v² = u² + 2as with v=0, a = −g gives H = u²/(2g).

Relative velocity of A with respect to B is the vector difference vA − vB.

When moving towards each other, relative speed = sum of magnitudes = 5 + 3 = 8 m/s.

Downstream speed = speed in still water + current = 6 + 2 = 8 km/h.

Relative speed = 36+54 = 90 km/h = 90×5/18 = 25 m/s. Total length = 250 m. Time = 250/25 = 10 s. (Wait — calculation shows 10 s). Correct answer should be B: 10 s.

Equal distance in equal times indicates constant speed — acceleration is zero.

a = (v − u)/t = (9 − 5)/2 = 4/2 = 2 m/s².

Let boat speed w.r.t water = v, current = c. v − c = 5, v + c = 9 → adding gives 2v = 14 ⇒ v = 7 ⇒ c = 2 km/h.

Relative speed in same direction = difference of speeds = 6 − 4 = 2 m/s.

v = u + at = 0 + 3×5 = 15 m/s.

a = (v − u)/t = (0 − 20)/4 = −20/4 = −5 m/s² (negative sign indicates deceleration).

s = ut + ½at² = 20×4 + 0.5×(−5)×16 = 80 − 40 = 40 m.

s = ut + ½at² → 200 = 0 + 0.5 a (10)² = 50 a → a = 200/50 = 4 m/s².

When time is not given and distance is needed, use v² = u² + 2as to solve for s (rearrange). If v unknown, choose equation containing knowns.

Always state/choose sign convention, ensure SI units (m, s), and select the kinematic equation containing your knowns and unknown — this prevents most mistakes.