JEE Physics Motion In A Straight Line MCQs Set 11

Answer: (c) \(5\pi R\), \(2R\)

Explanation:
1. 100 s is 2.5 rounds
2. Distance = \(2.5\times 2\pi R=5\pi R\)
3. After 2 full rounds he is back at the start; the half round takes him to the diametrically opposite point, so displacement = \(2R\)

Teacher's Note:
1. Distance counts the whole path length; displacement only compares end points.
2. Every full round adds distance but no displacement.

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

Explanation:
1. Displacement is the straight line between the two corners
2. Space diagonal = \(\sqrt{10^2+10^2+10^2}=10\sqrt{3}\) m

Teacher's Note:
1. Displacement does not depend on the path the fly takes.
2. The space diagonal of a cube of side a is \(a\sqrt3\).

Answer: (a) \(\dfrac{\pi}{180}\) m

Explanation:
1. The minute hand turns \(2\pi\) rad in 3600 s, so in 100 s it turns \(\theta=\dfrac{2\pi}{3600}\times100=\dfrac{\pi}{18}\) rad
2. Arc length = \(r\theta=0.1\times\dfrac{\pi}{18}=\dfrac{\pi}{180}\) m

Teacher's Note:
1. Arc length is radius times angle in radians.
2. Convert 10 cm to metres before comparing with the options.

Answer: (b) 50 m towards east

Explanation:
1. The last leg has components \(30\sqrt2\cos45^\circ=30\) m east and 30 m south
2. North-south: \(+30-30=0\)
3. East-west: \(20+30=50\) m east
4. Displacement is 50 m towards east

Teacher's Note:
1. Resolve each leg into east and north components and add.
2. Distance walked is 30 + 20 + 42.4 m, but displacement is only 50 m.

Answer: (a) 50 m, 53° east of south

Explanation:
1. Turning right through 143° from east gives a heading of 143° clockwise from east
2. Components of the second leg: \(x=50\cos143^\circ=-40\) m, \(y=-50\sin143^\circ=-30\) m
3. Net position: \(x=80-40=40\) m east, \(y=-30\) m (south)
4. Magnitude \(\sqrt{40^2+30^2}=50\) m, angle with east \(=\tan^{-1}(30/40)=37^\circ\) south of east, which is 53° east of south

Teacher's Note:
1. Angle with east 37° means angle with south 53°.
2. Always resolve the second leg using the direction after the turn.

Answer: (c) 37 s

Explanation:
1. Every 8 s cycle gives a net gain of 2 m
2. After 4 cycles (32 s) he is 8 m from the start
3. The next 5 forward steps take him to 13 m at 32 + 5 = 37 s, so he falls into the pit at 37 s

Teacher's Note:
1. Check at which point the forward run first reaches the pit rather than only the net gain.
2. After 3 cycles he is only at 6 m, and 6 + 5 = 11 m is short of the pit.

Answer: (c) 42 km/h

Explanation:
1. For equal distances, average speed is the harmonic mean
2. \(v_{av}=\dfrac{2v_1v_2}{v_1+v_2}=\dfrac{2\times30\times70}{100}=42\) km/h

Teacher's Note:
1. Average speed is total distance divided by total time, not the mean of the speeds.
2. Equal distances give the harmonic mean; equal times give the arithmetic mean.

Answer: (b) In general, average speed is greater than or equal to the magnitude of average velocity

Explanation:
1. Average velocity = displacement/time, so (a) is false
2. Path length is never smaller than the magnitude of displacement, so average speed \(\ge|\)average velocity\(|\), which makes (b) true
3. Speed is the magnitude of velocity, so non-zero velocity means non-zero speed; (c) is false
4. Equality in (d) holds only for motion along a straight line without reversal, so it is not true in general

Teacher's Note:
1. Distance is always greater than or equal to the magnitude of displacement.
2. Equality holds only for straight-line motion in one direction.

Answer: (b) 5 m/s

Explanation:
1. East displacement = 30 m, north displacement = 40 m
2. Net displacement = \(\sqrt{30^2+40^2}=50\) m
3. Average velocity = 50/10 = 5 m/s

Teacher's Note:
1. Average velocity is net displacement divided by total time.
2. Add perpendicular displacements as vectors, not as plain numbers.

Answer: (b) 2

Explanation:
1. Let each side be s
2. Distance = 2s, displacement (from A to C) = s
3. With the same time for both, the ratio = 2s/s = 2

Teacher's Note:
1. Average speed over magnitude of average velocity equals distance over displacement.
2. This ratio can never be less than 1.

Answer: (c) 66.7 m/s

Explanation:
1. \(v_{av}=\dfrac{2v_1v_2}{v_1+v_2}=\dfrac{2\times100\times50}{150}\)
2. \(=66.7\) m/s

Teacher's Note:
1. Equal distances at two speeds always give the harmonic mean.
2. The arithmetic mean 75 m/s is the common wrong answer.

Answer: (a) \(\dfrac{v_1+v_2+v_3}{3}\)

Explanation:
1. For total time T the distance is \(\dfrac T3(v_1+v_2+v_3)\)
2. Average velocity = distance/time \(=\dfrac{v_1+v_2+v_3}{3}\)

Teacher's Note:
1. Equal time intervals give the arithmetic mean of the velocities.
2. Equal distances would give the harmonic mean instead.

Answer: (d) \(\dfrac{5v_1v_2}{3v_1+2v_2}\)

Explanation:
1. Time \(=\dfrac{2d/5}{v_1}+\dfrac{3d/5}{v_2}=\dfrac{d(2v_2+3v_1)}{5v_1v_2}\)
2. Average speed \(=\dfrac dt=\dfrac{5v_1v_2}{3v_1+2v_2}\)

Teacher's Note:
1. Weight each speed by the fraction of distance it covers.
2. Check the limit \(v_1=v_2=v\): the answer reduces to v.

Answer: (d) 7.5 m/s

Explanation:
1. In the second half, equal times mean the average speed is \((4.5+7.5)/2=6\) m/s
2. For the whole journey, equal distances at 10 and 6 m/s: \(\dfrac{2\times10\times6}{10+6}=7.5\) m/s

Teacher's Note:
1. Break the journey into halves and treat each half by its own average speed.
2. Then combine using equal distances.

Answer: (d) 14.14 m/s in south-west direction

Explanation:
1. Initial velocity \(=10\hat i\), final velocity (due south) \(=-10\hat j\)
2. \(\Delta\vec v=\vec v_f-\vec v_i=-10\hat i-10\hat j\)
3. Magnitude \(=10\sqrt2=14.14\) m/s, pointing south-west

Teacher's Note:
1. Velocity is a vector, so a change of direction is a change of velocity even at constant speed.
2. Subtract the initial vector from the final vector.

Answer: (c) 6 cm/s²

Explanation:
1. \(v=2bt\) and \(a_{cc}=2b\)
2. \(2\times3=6\) cm/s², independent of time

Teacher's Note:
1. A quadratic position function gives constant acceleration.
2. The constant a = 15 cm only fixes the starting position.

Answer: (d) 50 m/s²

Explanation:
1. \(v=9t^2+14t+14\)
2. \(a=18t+14\)
3. At t = 2: \(a=36+14=50\) m/s²

Teacher's Note:
1. Differentiate twice for acceleration.
2. Substitute the time only after finishing the differentiation.

Answer: (a) 2 m/s²

Explanation:
1. \(x=(t-3)^2\)
2. \(v=2(t-3)\), \(a=2\) m/s²

Teacher's Note:
1. Rearrange the relation to get x as a function of t before differentiating.
2. The acceleration is constant.

Answer: (c) the acceleration of the particle is 2a

Explanation:
1. \(v=u+2a(t-2)\) and \(\dfrac{dv}{dt}=2a\), so (c) is true and (b) false
2. Initial velocity (t = 0) is \(u-4a\), not u
3. At t = 2, \(x=0\), so the particle is at the origin

Teacher's Note:
1. Check each option by substituting t = 0 and t = 2.
2. The shift (t - 2) changes the initial values but not the acceleration.

Answer: (a) 2 s

Explanation:
1. \(v=12t-3t^2\)
2. \(a=12-6t\)
3. \(a=0\) gives \(t=2\) s

Teacher's Note:
1. Zero acceleration is the moment of maximum velocity here.
2. Do not confuse it with the instant when velocity is zero (t = 4 s).

Answer: (d) -6 m

Explanation:
1. \(v=3t^2-12t+3\), \(a=6t-12\)
2. \(a=0\) at t = 2 s
3. \(s=8-24+6+4=-6\) m

Teacher's Note:
1. First find the time at which acceleration vanishes, then the displacement.
2. A negative displacement just means the particle is on the negative side of the origin.