JEE Physics Motion In A Straight Line MCQs Set 03

Answer: (c) 55 m

Explanation:
1. 0 to 1 s: \( \frac12\times1\times20=10 \) m. 1 to 2 s: \( 20\times1=20 \) m.
2. 2 to 3 s: \( \frac{20+10}2\times1=15 \) m. 3 to 4 s: \( 10\times1=10 \) m.
3. Total \( =55 \) m.

Teacher's Note:
1. Add the areas of the triangle, rectangle and trapezia.

Answer: (b) 4 s

Explanation:
1. \( x=\frac g2 \). The distance in the nth second is \( \frac g2(2n-1) \).
2. \( \frac g2(2n-1)=7\cdot\frac g2\Rightarrow n=4 \) s.

Teacher's Note:
1. The distances in successive seconds are in the ratio 1 : 3 : 5 : 7.

Answer: (a) \( v\propto t \)

Explanation:
1. \( \frac{dx}{dt}=\alpha\sqrt x\Rightarrow\int\frac{dx}{\sqrt x}=\alpha t\Rightarrow2\sqrt x=\alpha t \).
2. \( x=\frac{\alpha^2t^2}4 \), so \( v=\alpha\sqrt x=\frac{\alpha^2t}2\propto t \).

Teacher's Note:
1. Separate the variables and integrate.

Answer: (c) \( x=9t^2+8 \)

Explanation:
1. For constant acceleration x must be a quadratic polynomial in t.
2. Only \( x=9t^2+8 \) is quadratic, with \( a=18 \) m/s\( ^2 \).

Teacher's Note:
1. Differentiate twice.

Answer: (c) in the ratio of the differences of the square roots of the integers, i.e. \( \sqrt1,(\sqrt2-\sqrt1),(\sqrt3-\sqrt2),\dots \)

Explanation:
1. Time to fall n km from rest is \( t_n\propto\sqrt n \).
2. The time for the nth kilometre is \( t_n-t_{n-1}\propto\sqrt n-\sqrt{n-1} \).

Teacher's Note:
1. The speed keeps increasing, so later kilometres take less time.

Answer: (a) More than 19.6 m/s

Explanation:
1. A ball stays up for \( \frac{2u}g \). For more than two balls in the air at once, this must exceed \( 2\times2=4 \) s.
2. \( \frac{2u}g>4\Rightarrow u>19.6 \) m/s.

Teacher's Note:
1. Compare the time of flight with the spacing of throws.

Answer: (c) \( \frac43 \)

Explanation:
1. Time average over the motion to final speed v: \( \frac v2 \).
2. Space average: \( \frac{\int v\,ds}S \) with \( ds=\frac{v\,dv}a \) and \( S=\frac{v^2}{2a} \): \( \frac{v^3/3a}{v^2/2a}=\frac{2v}3 \).
3. Ratio \( =\frac{2v/3}{v/2}=\frac43 \).

Teacher's Note:
1. Average over distance weights the faster parts more.

Answer: (a) 8.3 m

Explanation:
1. Average velocities: \( \frac{10}5=2 \) m/s at \( t=2.5 \) s and \( \frac{10}3 \) m/s at \( t=6.5 \) s.
2. \( a=\frac{10/3-2}{4}=\frac13 \) m/s\( ^2 \). Velocity at \( t=9 \) s \( =2+\frac13\times6.5=\frac{25}6 \) m/s.
3. Distance in the next 2 s (8 s to 10 s) \( =\frac{25}6\times2\approx8.3 \) m.

Teacher's Note:
1. The average velocity equals the instantaneous velocity at the mid time.

Answer: (a) 2 m/s

Explanation:
1. Displacement \( =40-60=-20 \) m.
2. Average velocity \( =\frac{-20}{10}=-2 \) m/s, magnitude 2 m/s.

Teacher's Note:
1. Average velocity depends only on the end positions.

Answer: (a) Both cannot be zero

Explanation:
1. A change of location means the final position differs from the initial one, so the displacement is not zero.
2. The distance is at least as large as the displacement, so it is not zero either.
3. Hence both cannot be zero.

Teacher's Note:
1. Distance is never negative.

Answer: (c) 60 s

Explanation:
1. At burn-out \( v=10\times30=300 \) m/s.
2. After this the rocket decelerates at g: time to stop \( =\frac{300}{10}=30 \) s.
3. Total time \( =30+30=60 \) s.

Teacher's Note:
1. Treat the powered and free flight stages separately.

Answer: (b) 20 s

Explanation:
1. Time of flight \( =\frac{2u}g=\frac{200}{10}=20 \) s.

Teacher's Note:
1. Time up equals time down.

Answer: (b) 4 m/s

Explanation:
1. Equal time intervals: average velocity \( =\frac{3+4+5}3=4 \) m/s.

Teacher's Note:
1. With equal times, the average velocity is the simple mean.

Answer: (c) S1, S3 and S4

Explanation:
1. S1 is wrong: average velocity is displacement divided by time.
2. S2 is correct: distance is at least the magnitude of displacement.
3. S3 is wrong: a non-zero velocity means a non-zero speed.
4. S4 is wrong in general: it holds only for straight line motion without reversal.

Teacher's Note:
1. Define each quantity carefully.

Answer: (b) \( -a\cos t \)

Explanation:
1. \( v=\frac{dx}{dt}=-a\sin t \).
2. \( a_{cc}=\frac{dv}{dt}=-a\cos t \).

Teacher's Note:
1. Differentiate twice.

Answer: (b) the time of ascent is less than the time of descent

Explanation:
1. During ascent both gravity and air resistance act downwards: \( a_{up}=g+f/m \).
2. During descent air resistance acts upwards: \( a_{down}=g-f/m \).
3. Since the distance is the same and \( a_{up}>a_{down} \), the ascent takes less time.

Teacher's Note:
1. Compare the magnitudes of acceleration on the way up and down.

Answer: (c) 610 m

Explanation:
1. The stone starts with 12 m/s downwards: \( s=12\times10+\frac12(9.8)(100)=120+490=610 \) m.

Teacher's Note:
1. The initial velocity equals the balloon's velocity.

Answer: (b) 0, 10 m/s

Explanation:
1. Over one lap the displacement is zero, so the average velocity is 0.
2. Average speed \( =\frac{2\pi\times100}{62.8}=10 \) m/s.

Teacher's Note:
1. Average velocity uses displacement; average speed uses distance.

Answer: (c) \( 1:1:1 \)

Explanation:
1. By energy conservation \( \frac12mv^2=mgh \), so \( v=\sqrt{2gh} \).
2. It is independent of mass and path, so the ratio is 1 : 1 : 1.

Teacher's Note:
1. Frictionless paths conserve mechanical energy.

Answer: (d) 12 s

Explanation:
1. At the meeting time, \( 98t-4.9t^2=98(t-4)-4.9(t-4)^2 \).
2. Simplifying: \( 0=-392+39.2t-78.4\Rightarrow t=\frac{470.4}{39.2}=12 \) s.

Teacher's Note:
1. Equate the heights of the two bodies.

Answer: (c) reach the ground in 4 s

Explanation:
1. At release: \( v=1.25\times8=10 \) m/s upward and \( h=\frac12(1.25)(64)=40 \) m.
2. For the stone: \( -40=10t-5t^2\Rightarrow t^2-2t-8=0\Rightarrow t=4 \) s.
3. Its displacement is 40 m (downward), it first rises 5 m so the distance is 50 m, and it first moves upward.

Teacher's Note:
1. The stone leaves with the balloon's velocity.

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

Explanation:
1. Free fall from rest gives \( s_n\propto(2n-1) \), so the assertion is true.
2. Zero speed means zero velocity, so the reason is false.

Teacher's Note:
1. Speed is the magnitude of velocity.

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

Explanation:
1. A body at rest in one frame can be moving in another frame, so rest is relative.
2. All motion is relative to a frame of reference; there is no absolute motion, so the reason is false.

Teacher's Note:
1. Motion needs a frame of reference.