Uniformly Accelerated Motion

Answer: D. From rest, the distance in the nth second is proportional to (2n − 1). For n = 1, 2, 3: the ratios are 1 : 3 : 5. This follows from the second kinematic equation x = ½gt² (NCERT Class 11 Physics Chapter 2, page 17); the difference between successive cumulative distances gives the odd-number series.

Why A is wrong: A is wrong because 1:2:3 assumes distance grows linearly with time — that's the arithmetic-progression error. Distance grows as t² under constant acceleration, so successive-interval distances follow the odd-number rule, not a linear sequence (trap: free-fall linear vs quadratic).

Why B is wrong: B is wrong because equal distances in each second would require constant velocity, not constant acceleration. Under constant acceleration from rest, velocity increases with time, so later seconds cover more distance.

Why C is wrong: C is wrong because 1:4:9 gives the ratio of cumulative distances from t = 0 to t = 1, 2, 3 s (i.e., total distance ∝ t²), not the distance covered in each individual second. The question asks for per-second intervals, which follow the (2n−1) pattern.

Answer: A. Using v² = u² + 2gh with u = 10 m/s, v = 30 m/s, g = 10 m/s²: 900 = 100 + 2(10)h → h = 800/20 = 40 m. The initial velocity term u² is essential because the ball is thrown, not dropped (NCERT Class 11 Physics Chapter 2, page 18).

Why B is wrong: B is wrong because 20 m results from using v = u + gt to find t = 2 s and then applying h = ½gt² = ½(10)(4) = 20 m — this omits the initial velocity contribution to displacement. The correct displacement formula is h = ut + ½gt², or equivalently v² = u² + 2gh (trap: nonzero initial velocity dropped).

Why C is wrong: C is wrong because 45 m comes from using v² = 2gh (dropping the u² term): 900 = 2(10)h gives h = 45 m. This treats the ball as dropped from rest, ignoring the given initial speed of 10 m/s (trap: nonzero initial velocity dropped).

Why D is wrong: D is wrong because 80 m comes from dropping the factor 2 in v² = u² + 2gh: (900 − 100)/10 = 80 m. The 2 in 2gh halves the result to 40 m.

Answer: C. Uniformly accelerated motion means the acceleration remains constant — both its magnitude and direction do not change with time (NCERT Class 11 Physics Chapter 2, page 15). This is the condition under which the three kinematic equations are valid.

Why A is wrong: A is wrong because constant velocity means zero acceleration, which is uniform motion, not uniformly accelerated motion. These are distinct concepts.

Why B is wrong: B is wrong because if acceleration itself increases with time, the motion is non-uniformly accelerated. 'Uniform acceleration' means the acceleration value stays the same, not that it grows steadily.

Why D is wrong: D is wrong because displacement proportional to time (x ∝ t) implies constant velocity and zero acceleration. Under constant acceleration from rest, displacement is proportional to t², not t.

Answer: D. From rest with constant acceleration a, total distance at time t is s = ½at². At t = 1, 2, 3: s₁ : s₂ : s₃ = 1² : 2² : 3² = 1 : 4 : 9 (NCERT Class 11 Physics Chapter 2, page 17).

Why A is wrong: A is wrong because 1:3:5 gives the ratio of distances in successive individual seconds (the odd-number rule), not the cumulative distance from the start. The question asks for total distance at the end of each second.

Why B is wrong: B is wrong because 1:2:3 implies displacement grows linearly with time, which holds only for constant velocity. Under constant acceleration, displacement grows as t² (trap: free-fall linear vs quadratic).

Why C is wrong: C is wrong because 1:8:27 would correspond to s ∝ t³, which does not follow from constant acceleration. The second kinematic equation gives s ∝ t² for motion from rest.

Answer: B. The kinematic equations are derived under the assumption that acceleration is constant in magnitude and direction (NCERT Class 11 Physics Chapter 2, page 19). They fail for variable acceleration or curvilinear motion where acceleration direction changes.

Why A is wrong: A is wrong because constant speed along a curved path means the velocity direction changes, so acceleration (centripetal) is present and not along the direction of motion. The kinematic equations apply to straight-line constant-acceleration motion, not curved paths.

Why C is wrong: C is wrong because constant velocity means zero acceleration, so s = vt is sufficient. The kinematic equations are trivially satisfied but are designed for the non-trivial case where acceleration is nonzero and constant.

Why D is wrong: D is wrong because linearly varying acceleration (a = a₀ + bt) violates the constant-acceleration condition. For such cases, you must integrate a(t) to find v(t) and then x(t) (mistake: applying kinematic equations to non-uniform acceleration).

Answer: B. Differentiating t = αx² + βx with respect to x: dt/dx = 2αx + β, so v = dx/dt = 1/(2αx + β). For acceleration: a = dv/dt = (dv/dx)(dx/dt) = v(dv/dx). From v = (2αx + β)⁻¹: dv/dx = −2α(2αx + β)⁻² = −2αv². Therefore a = v × (−2αv²) = −2αv³.

Why A is wrong: A is wrong because −2αv² is the intermediate result dv/dx, not the acceleration. The acceleration requires one more multiplication by v: a = v(dv/dx) = v(−2αv²) = −2αv³. Stopping at dv/dx is a common chain-rule omission (trap: implicit differentiation kinematics).

Why C is wrong: C is wrong because the sign is positive and the power is v² instead of v³. The derivative dv/dx is negative (since increasing x increases dt/dx, which decreases v), and a = v × dv/dx introduces one additional factor of v, giving −2αv³.

Why D is wrong: D is wrong because −αv² has the wrong coefficient (missing the factor of 2 from the derivative of αx²) and the wrong power of v. The correct chain-rule application yields −2αv³.

Answer: A. Distance in the nth second from rest: sₙ = ½g(2n − 1). For n = 4: s₄ = ½ × 10 × (2 × 4 − 1) = 5 × 7 = 35 m. This follows from the difference s(4) − s(3) = ½g(16 − 9) = ½g(7) (NCERT Class 11 Physics Chapter 2, page 17).

Why B is wrong: B is wrong because 25 m could result from computing ½g(n) = ½(10)(5) with n = 5 by miscounting, or from confusing the formula. The correct expression uses (2n − 1) = 7, not 5.

Why C is wrong: C is wrong because 40 m results from using sₙ = ½g(2n) = ½(10)(8) = 40 m — inserting 2n instead of (2n − 1). The odd-number factor must be (2n − 1), which equals 7 for n = 4.

Why D is wrong: D is wrong because 80 m is the total cumulative distance fallen in 4 s: s = ½g t² = ½(10)(16) = 80 m. The question asks for the distance in the 4th second alone, not the total.

Answer: C. At maximum height, v = 0. Using v² = u² − 2gh (upward throw, g opposes motion): 0 = (20)² − 2(10)h → h = 400/20 = 20 m (NCERT Class 11 Physics Chapter 2, page 18).

Why A is wrong: A is wrong because 10 m results from using h = u/2g = 20/20 = 1 m (dimensionally wrong) or h = u²/4g = 400/40 = 10 m, which introduces an extra factor of 2 in the denominator. The correct formula is h = u²/(2g).

Why B is wrong: B is wrong because 40 m results from using h = u²/g = 400/10 = 40 m, omitting the factor of 2 in the denominator. The third kinematic equation gives v² = u² − 2gh, so h = u²/(2g), not u²/g.

Why D is wrong: D is wrong because 80 m results from using h = 2u²/g = 2(400)/10 = 80 m, which inverts the ½ factor. The correct relationship h = u²/(2g) = 20 m.