Motion — Set 4
Physics · गति · Questions 31–40 of 60
Which graph represents uniform motion (constant velocity) in a distance–time plot?
Correct Answer: A. A straight line with constant slope
• **A straight line with constant slope** = In a distance–time graph, slope = speed; constant speed produces a straight line with a fixed positive slope. • **constant slope = constant speed** — the steeper the line, the higher the uniform speed; a horizontal line means zero speed. • 💡 Wrong-option analysis: A curved line with increasing slope: this represents uniformly accelerated motion (increasing speed); A horizontal line only: a horizontal line means the body is at rest (zero speed); A zig-zag line: this represents erratic, non-uniform motion changing direction repeatedly.
A runner completes 400 m in 50 s. What is the average speed?
Correct Answer: A. 8 m/s
• **8 m/s** = Average speed = distance / time = 400 m / 50 s = 8 m/s. • **400 ÷ 50 = 8 m/s** — straightforward division of total distance by total time. • 💡 Wrong-option analysis: 6 m/s: 6×50 = 300 m, not 400 m; 9 m/s: 9×50 = 450 m, not 400 m; 7 m/s: 7×50 = 350 m, not 400 m.
Instantaneous speed is the speed?
Correct Answer: A. At a particular instant of time
• **At a particular instant of time** = Instantaneous speed is the magnitude of the instantaneous velocity — the speed at one specific moment. • **|v| at a single instant** — mathematically it is the limit of Δd/Δt as Δt → 0; a car's speedometer reads instantaneous speed. • 💡 Wrong-option analysis: Only when acceleration is zero: instantaneous speed is defined at every instant regardless of acceleration; Only in circular motion: it applies to all types of motion; Over a full journey: that describes average speed, not instantaneous speed.
A wheel rotates at 120 rpm. What is its angular speed in rad/s? (Take π ≈ 3.14)?
Correct Answer: C. 4π rad/s
• **4π rad/s** = 120 rpm = 120/60 = 2 rev/s; each revolution = 2π rad, so ω = 2 × 2π = 4π rad/s. • **ω = 2π × (rpm/60) = 2π × 2 = 4π rad/s** — always convert rpm to rev/s first, then multiply by 2π. • 💡 Wrong-option analysis: 6π rad/s: would require 3 rev/s = 180 rpm, not 120 rpm; 2π rad/s: corresponds to 1 rev/s = 60 rpm, not 120 rpm; 8π rad/s: would require 4 rev/s = 240 rpm, not 120 rpm.
In uniform circular motion, the acceleration of a particle is directed?
Correct Answer: D. Along the radius towards the center
• **Along the radius towards the center** = In uniform circular motion, the centripetal acceleration is directed radially inward — always toward the center of the circular path. • **centripetal = center-seeking** — it changes the direction of velocity continuously without changing its magnitude. • 💡 Wrong-option analysis: Opposite to velocity always: velocity is tangential; 'opposite to velocity' would be tangential backward — not the centripetal direction; Along the radius away from the center: this is the centrifugal (outward pseudo-force) direction, not real acceleration; Along the tangent: the tangent is the direction of velocity, not acceleration.
A body moves with uniform acceleration from rest and reaches 12 m/s in 6 s. How much distance does it cover in this time?
Correct Answer: B. 36 m
• **36 m** = Average velocity = (u + v)/2 = (0 + 12)/2 = 6 m/s; distance = v_avg × t = 6 × 6 = 36 m. • **s = ½(u+v)t = ½×12×6 = 36 m** — also verifiable: a = 12/6 = 2 m/s², s = ½×2×36 = 36 m. • 💡 Wrong-option analysis: 24 m: arithmetic error not matching any correct formula with these values; 72 m: error of using s = v×t = 12×6 = 72 (using final velocity without the ½ factor); 48 m: error possibly from s = a×t² = 2×36 = 72? or a different arithmetic mistake.
Which equation is correct for displacement in uniformly accelerated motion?
Correct Answer: D. s = ut + (1/2)at^2
• **s = ut + ½at²** = This is the second equation of motion for constant acceleration; displacement = (initial velocity × time) + (½ × acceleration × time²). • **s = ut + ½at²** — derived by integrating v = u + at over time; for u = 0 it reduces to s = ½at². • 💡 Wrong-option analysis: s = vt: valid only at constant velocity (a = 0), not for accelerated motion; s = (u−v)/t: dimensionally wrong — this gives acceleration, not displacement; s = u + at: this is the formula for final velocity v, not displacement.
A body covers 100 m in 5 s starting from rest with uniform acceleration. What is the acceleration?
Correct Answer: A. 8 m/s^2
• **8 m/s²** = Using s = ½at² with u = 0: 100 = ½ × a × 25 → a = 200/25 = 8 m/s². • **a = 2s/t² = 2×100/25 = 8 m/s²** — starting from rest allows direct use of s = ½at². • 💡 Wrong-option analysis: 2 m/s²: gives s = ½×2×25 = 25 m, not 100 m; 10 m/s²: gives s = ½×10×25 = 125 m, not 100 m; 4 m/s²: gives s = ½×4×25 = 50 m, not 100 m.
In free fall from rest (no air resistance), the distances covered in successive equal time intervals are in the ratio?
Correct Answer: D. 1:3:5
• **1:3:5** = For uniformly accelerated motion from rest, distances in successive equal intervals follow Galileo's odd-number law: (2n−1) for nth interval gives 1, 3, 5, 7, … • **Galileo's odd-number law** — distance in nth second = u + a(2n−1)/2; for u = 0, it is proportional to (2n−1). • 💡 Wrong-option analysis: 2:4:6: even numbers — this incorrect ratio has no physical basis; 1:1:1: equal distances imply constant velocity, not acceleration; 1:2:3: linear ratio applies to velocities at equal times from rest, not distances.
A projectile is thrown with speed 20 m/s at 45°. Taking g = 10 m/s^2, what is its range?
Correct Answer: D. 40 m
• **40 m** = Range R = u² sin2θ/g; with u = 20 m/s, θ = 45°, g = 10 m/s²: R = 400 × sin90°/10 = 400/10 = 40 m. • **R = u²/g at 45° = 400/10 = 40 m** — sin(2×45°) = sin90° = 1 maximises range at this angle. • 💡 Wrong-option analysis: 50 m: would need u² = 500, i.e. u ≈ 22.4 m/s; 20 m: would need u² = 200, i.e. u ≈ 14.1 m/s; 30 m: would need u² = 300, i.e. u ≈ 17.3 m/s.