Motion in a Straight Line – Terms Related to Motion: NEET MCQs with Answers

Explanation: With equal distances covered at two different speeds, v_avg = 2v₁v₂/(v₁ + v₂) = 2(40)(60)/(40 + 60) = 4800/100 = 48 km/h. The answer is deliberately less than the arithmetic mean 50 km/h, because the car spends more time on the slower half of the journey.

Explanation: For two equal halves covered at different speeds, the average speed is the harmonic mean: v_avg = 2(40)(v)/(40 + v) Substituting v_avg = 48: 48 = 80v/(40 + v) 48(40 + v) = 80v 1920 + 48v = 80v 32v = 1920 v = 60 km/h. Note that the total distance of 200 m never enters the calculation, since the average speed for equal halves depends only on the two speeds.

Explanation: Let the total distance be s, so each stage is s/3. t₁ = (s/3)/10, t₂ = (s/3)/20, t₃ = (s/3)/60 Total time = (s/3)(1/10 + 1/20 + 1/60) = (s/3) × (6 + 3 + 1)/60 = (s/3)(10/60) = s/18 Average speed = s/(s/18) = 18 km/h. As always with equal distances, the result is the harmonic mean of the three speeds — far closer to the smallest speed than to the largest.

Explanation: Velocity, v = ds/dt = 3t² − 12t + 3 …(i) Acceleration, a = dv/dt = 6t − 12 Setting a = 0: 6t − 12 = 0 ⇒ t = 2 s Substituting t = 2 s into (i): v = 3(2)² − 12(2) + 3 = 12 − 24 + 3 = −9 ms⁻¹. The negative sign shows that at this instant the particle is moving in the direction opposite to the positive s-direction.

Explanation: Let t₁ be the time of acceleration and t₂ the time of deceleration, with t₁ + t₂ = t. Let v_max be the velocity reached at the end of the accelerating phase. For the accelerating phase: v_max = αt₁ ⇒ t₁ = v_max/α For the decelerating phase (ending at rest): 0 = v_max − βt₂ ⇒ t₂ = v_max/β Adding: t = v_max/α + v_max/β = v_max(α + β)/(αβ) Therefore v_max = αβt/(α + β). On a v–t graph this situation is a triangle of base t and height v_max, with the two sides having slopes α and −β.

Explanation: Since the acceleration is not constant, the standard equations of motion cannot be used; we must integrate. Given a = dv/dt = bt. Integrating once: v = ∫bt dt = bt²/2 + C₁. At t = 0, v = v₀, so C₁ = v₀ and v = v₀ + bt²/2 Integrating again, since v = dx/dt: x = ∫(v₀ + bt²/2) dt = v₀t + bt³/6 + C₂ The particle starts from the origin, so x = 0 at t = 0 giving C₂ = 0. Hence x = v₀t + (1/6)bt³.

Explanation: Velocity, v = dx/dt = 2at − 3bt² Acceleration, A = dv/dt = 2a − 6bt Setting the acceleration equal to zero: 2a − 6bt = 0 ⇒ 6bt = 2a ⇒ t = 2a/6b = a/3b. Beyond this instant the acceleration becomes negative, so t = a/3b is also the instant of maximum velocity.

Explanation: Velocity, v = ds/dt = 9t² + 14t + 14 Acceleration, a = dv/dt = 18t + 14 At t = 1 s: a = 18(1) + 14 = 32 m/s². Note that the constant term 5 and the linear term 14t disappear on the second differentiation, so they do not affect the acceleration at all.

Explanation: Velocity, v = dx/dt = −aαe⁻ᵅᵗ + bβe^(βt) Differentiating once more to get the acceleration: dv/dt = aα²e⁻ᵅᵗ + bβ²e^(βt) Since a, b, α and β are all positive and an exponential function is always positive, both terms on the right are positive. Hence dv/dt > 0 at every instant. A permanently positive acceleration means the velocity keeps on increasing with time.

Explanation: After one complete lap the car returns to its starting point, so its displacement is zero. Average velocity = displacement/time = 0/62.8 = 0 The distance covered in one lap is the circumference: d = 2πr = 2 × 3.14 × 100 = 628 m Average speed = distance/time = 628/62.8 = 10 m/s. This question illustrates the key distinction between the two quantities: average velocity depends on displacement (a vector), average speed on the actual path length (a scalar).

Explanation: Velocity, v = dx/dt = 12 − 3t² The particle comes to rest when v = 0: 12 − 3t² = 0 ⇒ t² = 4 ⇒ t = 2 s Since v = 12 − 3t² is positive for every t < 2 s, the particle moves steadily in one direction and never reverses during this interval. The distance travelled is therefore simply the change in position. Position at t = 0: x₀ = 40 + 0 − 0 = 40 m Position at t = 2 s: x = 40 + 12(2) − (2)³ = 40 + 24 − 8 = 56 m Distance travelled = 56 − 40 = 16 m. The constant term 40 m is merely the particle's initial distance from O; it is not part of the distance covered during the motion.

Explanation: Velocity, v = dx/dt = 18t − 3t² The speed is maximum when its rate of change — the acceleration — becomes zero: a = dv/dt = 18 − 6t = 0 ⇒ t = 3 s Substituting t = 3 s into the position equation: x = 9(3)² − (3)³ = 9(9) − 27 = 81 − 27 = 54 m.

Explanation: Let XY = d, so the total distance for the round trip is 2d. t₁ = d/v_u (outward) and t₂ = d/v_d (return) Total time = d/v_u + d/v_d = d(v_u + v_d)/(v_u v_d) Average speed = 2d ÷ [d(v_u + v_d)/(v_u v_d)] = 2v_u v_d/(v_u + v_d). Note that the average velocity for the round trip is zero, since the net displacement is zero — average speed and average velocity are quite different here.

Explanation: First find when the acceleration vanishes: f = f₀(1 − t/T) = 0 ⇒ 1 − t/T = 0 ⇒ t = T (since f₀ ≠ 0) Now, since f = dv/dt, integrate from t = 0 (where v = 0) to t = T: v = ∫₀ᵀ f₀(1 − t/T) dt = f₀[t − t²/2T]₀ᵀ = f₀(T − T²/2T) = f₀(T − T/2) = ½f₀T.

Explanation: Given x = (t + 5)⁻¹. Velocity, v = dx/dt = −(t + 5)⁻² …(i) Acceleration, a = dv/dt = 2(t + 5)⁻³ …(ii) From (i), taking magnitudes, |v| = (t + 5)⁻², so |v|^(3/2) = [(t + 5)⁻²]^(3/2) = (t + 5)⁻³ Comparing with (ii), a = 2|v|^(3/2), i.e. a ∝ v^(3/2).