Motion — Set 3

Correct Answer: C. Velocity

• **Velocity** = Velocity specifies both the magnitude (speed) and direction of motion, making it a vector quantity. • **vector = magnitude + direction** — changing only the direction of motion changes velocity but not speed (e.g., uniform circular motion). • 💡 Wrong-option analysis: Mass: mass is a scalar — it has only magnitude and no direction; Time: time is a scalar — it flows in one direction and has no spatial direction; Speed: speed is the magnitude of velocity, a scalar with no direction information.

Correct Answer: A. 2.5 m/s

• **2.5 m/s** = 100 cm = 1 m, so divide by 100: 250 cm/s ÷ 100 = 2.5 m/s. • **÷ 100** — to convert cm to m always divide by 100; 250 ÷ 100 = 2.5. • 💡 Wrong-option analysis: 250 m/s: same number but wrong unit — 1 cm/s ≠ 1 m/s; 25 m/s: error of dividing by 10 instead of 100; 0.25 m/s: error of dividing by 1000 (the mm-to-m conversion factor).

Correct Answer: B. Air resistance equals its weight

• **Air resistance equals its weight** = Terminal velocity is reached when the upward air resistance exactly balances the downward gravitational force (weight), so net force = 0 and acceleration = 0. • **F_drag = mg → net force = 0** — at this point the object falls at constant (terminal) velocity with zero acceleration. • 💡 Wrong-option analysis: The object stops moving: wrong — at terminal velocity the object moves at constant nonzero speed; Its weight becomes zero: impossible — weight depends on mass and g, which remain constant; Acceleration becomes maximum: wrong — acceleration is maximum at the start of fall (before drag builds up) and becomes zero at terminal velocity.

Correct Answer: C. -2 m/s^2

• **-3 m/s²** = Using v² = u² + 2as: 10² = 20² + 2×a×50 → 100 = 400 + 100a → a = −300/100 = −3 m/s². • **a = (v²−u²)/(2s) = (100−400)/100 = −3 m/s²** — negative confirms deceleration. • 💡 Wrong-option analysis: −1.5 m/s²: arithmetic error in the formula; −4 m/s²: error of using a = (v−u)/s without squaring; −2 m/s²: would require s = 75 m, not 50 m.

Correct Answer: A. Four times

• **Four times** = Stopping distance s = u²/(2a); if speed doubles, u² becomes 4u², so stopping distance becomes four times larger. • **s ∝ u²** — kinetic energy (½mv²) is what the brakes must remove; doubling v quadruples KE and hence stopping distance. • 💡 Wrong-option analysis: Unchanged: wrong — stopping distance varies strongly with speed squared; Half: wrong — halving applies if speed were reduced to 1/√2 of original; Double: wrong — doubling would apply for a linear relationship, but KE and stopping distance scale as v².

Correct Answer: D. 40 km/h

• **40 km/h** = Average speed = total distance / total time = 120 km / 3 h = 40 km/h. • **120/3 = 40 km/h** — direct application of the definition of average speed. • 💡 Wrong-option analysis: 50 km/h: 50×3 = 150 km, not 120 km; 30 km/h: 30×3 = 90 km, not 120 km; 60 km/h: 60×3 = 180 km, not 120 km.

Correct Answer: A. Uniform straight-line motion

• **Uniform straight-line motion** = When a body moves in a straight line at constant speed, neither the magnitude nor direction of velocity changes, so acceleration = Δv/Δt = 0. • **a = 0, v ≠ 0** — Newton's first law: no net force means no acceleration and the body maintains its constant velocity. • 💡 Wrong-option analysis: Free fall: g ≈ 9.8 m/s² downward — acceleration is nonzero; Uniform circular motion: direction changes continuously, so centripetal acceleration is nonzero; Rest: both velocity and acceleration are zero — velocity is not nonzero here.

Correct Answer: D. 2 s

• **2 s** = Using s = ½gt² with s = 20 m, g = 10 m/s²: 20 = 5t² → t² = 4 → t = 2 s. • **t = √(2s/g) = √(40/10) = √4 = 2 s** — free fall from rest; the body is dropped (u = 0). • 💡 Wrong-option analysis: 1 s: gives s = 5×1 = 5 m, not 20 m; 3 s: gives s = 5×9 = 45 m, not 20 m; 4 s: gives s = 5×16 = 80 m, not 20 m.

Correct Answer: B. Negative acceleration

• **Negative acceleration** = Retardation means the body slows down; the acceleration vector points opposite to the direction of motion, which is negative in one-dimensional analysis. • **retardation = −a** — if a car decelerates at 5 m/s², its acceleration is −5 m/s² in the direction of motion. • 💡 Wrong-option analysis: Positive acceleration: positive acceleration means speeding up — the opposite of retardation; Zero displacement: zero displacement means the body has not moved — unrelated to retardation; Zero speed: zero speed is the final result after retardation stops the body, not the definition.

Correct Answer: B. 20 km/h

• **20 km/h** = For same-direction motion, relative speed = |v₁ − v₂| = |60 − 40| = 20 km/h. • **v_rel (same direction) = |v₁ − v₂| = 20 km/h** — from the slower car, the faster one pulls away at 20 km/h. • 💡 Wrong-option analysis: 100 km/h: this is the sum used for opposite-direction motion; 60 km/h: this is the faster car's absolute speed, not relative; 40 km/h: this is the slower car's absolute speed, not relative.