JEE Physics Motion In A Straight Line MCQs Set 19

Answer: (b) \( 10\sqrt{3} \) m

Explanation:
1. Displacement is the straight-line distance between the two corners, the body diagonal of the cube.
2. d = √(10² + 10² + 10²) = 10√3 m.

Teacher's Note:
1. Displacement ignores the path and depends only on end points.
2. The body diagonal of a cube of side a is a√3.

Answer: (b) 24 km/h

Explanation:
1. Let AB = d. Time = d/20 + d/30 = d/12.
2. Average speed = total distance / total time = 2d / (d/12) = 24 km/h.

Teacher's Note:
1. The average speed is not the arithmetic mean when equal distances are covered at different speeds.
2. Use the harmonic mean: 2v₁v₂/(v₁ + v₂).

Answer: (a) 3 m/s

Explanation:
1. Let the total distance be 3s. Times: s/2, s/3, s/6.
2. Total time = s(1/2 + 1/3 + 1/6) = s.
3. Average velocity = 3s / s = 3 m/s.

Teacher's Note:
1. Motion is in one direction, so average velocity equals average speed.
2. Add times, not speeds.

Answer: (b) velocity zero, speed 10 m/s

Explanation:
1. Distance in one lap = 2πr = 628 m, so average speed = 628/62.8 = 10 m/s.
2. After a full lap the car is back at its starting point: displacement zero, so average velocity is zero.

Teacher's Note:
1. Average velocity uses displacement, average speed uses path length.
2. A closed path always gives zero average velocity.

Answer: (a) gt

Explanation:
1. s = gt²/2.
2. v = ds/dt = gt.

Teacher's Note:
1. Velocity is the first derivative of displacement.
2. Differentiate t² to get 2t.

Answer: (c) 1 : 3 : 5

Explanation:
1. Distances from rest in 10, 20, 30 s are in the ratio 1² : 2² : 3² = 1 : 4 : 9.
2. Distances in successive equal intervals are 1 : (4 − 1) : (9 − 4) = 1 : 3 : 5.

Teacher's Note:
1. From rest, distances in successive equal intervals follow the odd numbers 1, 3, 5.
2. Take differences of the cumulative distances.

Answer: (a) \( v_T + v_B = 12 \) m/s

Explanation:
1. Average velocity across the window = 3/0.5 = 6 m/s.
2. For uniform acceleration this equals (v_T + v_B)/2, so v_T + v_B = 12 m/s.
3. Also v_B − v_T = g(0.5) = 4.9 m/s, so (b) as written (v_T − v_B) is negative and false.

Teacher's Note:
1. Average velocity under constant acceleration is the mean of initial and final velocities.
2. The ball speeds up, so v_B is greater than v_T.

Answer: (d) g downward, u

Explanation:
1. On release only gravity acts, so the acceleration is g downward.
2. Velocity cannot change instantly, so the stone still has speed u (upward).

Teacher's Note:
1. After release the lift's acceleration no longer matters to the stone.
2. Velocity is continuous; acceleration can jump.

Answer: (b) \( v = u + \dfrac{at^2}{2} \)

Explanation:
1. dv/dt = at.
2. Integrating from 0 to t gives v − u = at²/2.

Teacher's Note:
1. Acceleration that changes with time needs integration, not v = u + at.
2. ∫ t dt = t²/2.

Answer: (b) \( T = \sqrt{\dfrac{2h}{g}} + \dfrac{h}{v} \)

Explanation:
1. Fall time: h = ½gt₁², so t₁ = √(2h/g).
2. Sound returns the same distance h at speed v: t₂ = h/v.
3. T = t₁ + t₂.

Teacher's Note:
1. Two stages: free fall, then sound travelling up.
2. Add the times.

Answer: (b) \( \dfrac{275}{3} \) m/s

Explanation:
1. Student's fall time: 320 = 5t², so t = 8 s.
2. Superman has 8 − 5 = 3 s: 320 = 3u + ½(10)(3²) = 3u + 45.
3. u = 275/3 m/s.

Teacher's Note:
1. Superman must cover the same height in a shorter time.
2. Write both falls with s = ut + ½gt².

Answer: (c) 125 m

Explanation:
1. Even with an infinitely large initial speed Superman needs a non-zero start; if the student reaches the ground before 5 s he cannot be saved.
2. In 5 s the student falls ½(10)(25) = 125 m.
3. So the building height must be 125 m or less.

Teacher's Note:
1. Find the height dropped through before Superman even jumps.
2. Use h = ½gt² with t = 5 s.

Answer: (b) Car A always leads car B, and both cars move with the same velocity

Explanation:
1. Parallel lines have equal slopes, so the velocities are equal and neither car is faster.
2. Car A's line lies above car B's line at every time, so A is always ahead.
3. Straight lines mean constant velocity, so the acceleration is zero.

Teacher's Note:
1. Slope of an x-t graph is velocity.
2. Parallel lines mean equal velocity and zero acceleration.

Answer: (d) The particle moves at constant velocity up to a time t₀ and then stops

Explanation:
1. Up to t₀ the graph is a straight line with positive slope: constant positive velocity.
2. After t₀ the graph is horizontal: slope zero, so the particle is at rest.
3. The area under an x-t graph has no meaning as displacement.

Teacher's Note:
1. Read the slope, not the height.
2. A horizontal line on an x-t graph means rest.

Answer: (c) E

Explanation:
1. Velocity is the slope of the x-t graph.
2. At C the curve rises: positive. At D (the maximum) the slope is zero. At E the curve falls: negative slope. At F the curve rises again: positive.
3. So the velocity is negative at E.

Teacher's Note:
1. A maximum has zero slope.
2. Falling part of the curve gives negative velocity.

Answer: (c) 55 m

Explanation:
1. Distance = area under the v-t graph.
2. 0 to 1 s: triangle, ½ × 1 × 20 = 10 m.
3. 1 to 2 s: rectangle, 1 × 20 = 20 m.
4. 2 to 3 s: trapezium, ½(20 + 10) × 1 = 15 m.
5. 3 to 4 s: rectangle, 1 × 10 = 10 m.
6. Total = 10 + 20 + 15 + 10 = 55 m.

Teacher's Note:
1. Split the area into simple shapes.
2. The velocity never goes negative, so distance equals displacement.

Answer: (b) Graph (B)

Explanation:
1. v² = 2aS, so v = √(2aS).
2. v starts from zero and rises with S, but its slope decreases: a curve that bends towards the S axis, like a square root.
3. This is graph (B).

Teacher's Note:
1. Relate v and S with v² = u² + 2aS.
2. A square-root curve starts steep and flattens.

Answer: (d) \( \dfrac{1}{3} \)

Explanation:
1. Velocity = slope = tan θ.
2. v_A / v_B = tan 30° / tan 60° = (1/√3) / √3 = 1/3.

Teacher's Note:
1. The slope of the line, not the angle itself, gives the velocity.
2. tan 30° = 1/√3 and tan 60° = √3.

Answer: (d) A is false, but R is true.

Explanation:
1. Slowing down needs acceleration opposite to the velocity. A particle moving in the negative direction with negative acceleration speeds up, so A is false.
2. The direction of acceleration is set by the net force, independent of the velocity direction, so R is true.

Teacher's Note:
1. Compare the signs of velocity and acceleration, not the sign of acceleration alone.
2. A negative acceleration is not the same as retardation.

Answer: (d) A is false, but R is true.

Explanation:
1. At the top of a vertical throw v = 0 but a = g, so A is false.
2. The relation a = v dv/dx is a valid expression for acceleration, so R is true.

Teacher's Note:
1. Remember the top point of a thrown ball.
2. Zero velocity can coexist with non-zero acceleration.

Answer: (d) A is false, but R is true.

Explanation:
1. If u and v are equal and opposite (for example 10 m/s up and 10 m/s down in a vertical throw), (u + v)/2 = 0.
2. So the average velocity can be zero and A is false, while R is a true result.

Teacher's Note:
1. Constant acceleration can still reverse the direction of motion.
2. Use (u + v)/2 with signs.

Answer: (b) 75 m

Explanation:
1. First ball: y₁ = 40t − 5t². Second ball: y₂ = 40(t − 2) − 5(t − 2)².
2. Setting y₁ = y₂: 40t − 5t² = 40t − 80 − 5t² + 20t − 20, giving 20t = 100, t = 5 s.
3. y = 40(5) − 5(25) = 75 m.

Teacher's Note:
1. Measure both times from the first throw.
2. The t² terms cancel.

Answer: (b) 22 m/s²

Explanation:
1. x = ∫v dt = t⁴ − t². Set x = 2: t⁴ − t² − 2 = 0, so t² = 2.
2. a = dv/dt = 12t² − 2 = 12(2) − 2 = 22 m/s².

Teacher's Note:
1. Integrate velocity to get position, then solve for the time.
2. Use the positive root t² = 2.

Answer: (b) \( \left(\dfrac{\pi}{2\omega} - \dfrac{\theta}{\omega}\right) \)

Explanation:
1. x is maximum when sin(ωt + θ) = 1, i.e. ωt + θ = π/2.
2. t = (π/2 − θ)/ω.

Teacher's Note:
1. The first maximum of a sine occurs at π/2.
2. Solve the angle for t.