JEE Physics Motion In A Straight Line MCQs Set 24

Answer: (c) The retardation of the particle is \( \dfrac{1}{4(x+1)^3} \)

Explanation:
1. \( (x+1)^2=t+1 \), so \( x+1=\sqrt{t+1} \).
2. Velocity \( v=\frac{1}{2\sqrt{t+1}}=\frac{1}{2(x+1)} \) (positive, decreasing).
3. Acceleration \( a=\frac{dv}{dt}=-\frac{1}{4(t+1)^{3/2}}=-\frac{1}{4(x+1)^3} \): a retardation of magnitude \( \frac{1}{4(x+1)^3} \).

Teacher's Note:
1. Write x as a function of t first, then differentiate twice.
2. The negative acceleration with positive velocity means a retardation.

Answer: (b) 75 m

Explanation:
1. First ball after time \( t \): \( 40t-5t^2 \); second ball after \( t-2 \): \( 40(t-2)-5(t-2)^2 \).
2. Equating gives \( 20t=100 \), so \( t=5 \) s.
3. Height \( 200-125=75 \) m.

Teacher's Note:
1. Subtract the two height equations; the quadratic terms in t cancel, leaving a simple linear equation.

Answer: (a) \( \dfrac{3h}{4} \) metres from the ground

Explanation:
1. Distance dropped in \( T/2 \): \( \tfrac12g(T/2)^2=\frac14\cdot\frac12gT^2=\frac h4 \).
2. So the height above ground is \( h-\frac h4=\frac{3h}{4} \).

Teacher's Note:
1. Fall distance is proportional to time squared, so half the time gives a quarter of the height.

Answer: (d) \( \dfrac vg\left[1+\sqrt{1+\dfrac{2gh}{v^2}}\right] \)

Explanation:
1. Upward positive: \( -h=vt-\frac12gt^2 \), so \( gt^2-2vt-2h=0 \).
2. The positive root is \( t=\frac{v+\sqrt{v^2+2gh}}{g}=\frac vg\left[1+\sqrt{1+\frac{2gh}{v^2}}\right] \).

Teacher's Note:
1. Choose the root with the plus sign, since time cannot be negative.
2. The expression splits into time to rise and time to fall.

Answer: (c) 495 m

Explanation:
1. The stone starts with \( 10 \) m/s upward: \( -H=10(11)-\tfrac12(10)(11)^2 \).
2. So \( H=605-110=495 \) m.

Teacher's Note:
1. A dropped object shares the balloon's velocity at the moment of release, so it first keeps rising.

Answer: (b) \( \dfrac{15}{4}\ \text{m} \)

Explanation:
1. The first drop falls 5 m in \( 1 \) s, so the interval between drops is \( 0.5 \) s.
2. The second drop has been falling for \( 0.5 \) s: \( \tfrac12(10)(0.5)^2=1.25 \) m.
3. Its height \( 5-1.25=3.75=15/4 \) m.

Teacher's Note:
1. With three drops, the time from the first leaving to the third leaving is two intervals.

Answer: (a) (1), (2) and (4) only

Explanation:
1. \( a=10-t \) gives \( v=10t-\frac{t^2}{2} \): \( v(20)=0 \) and \( v>0 \) in between.
2. So the average acceleration is zero (1), the velocity never changes sign (2), the displacement is positive, not zero (3 is false).
3. The velocity curve is symmetric about \( t=10 \), so the average speeds in the two halves are equal (4).

Teacher's Note:
1. Integrate a to get v and check its sign.
2. A velocity that returns to zero without crossing zero means no turnaround.

Answer: (d) (2), (3) and (4) only

Explanation:
1. Zero acceleration at one instant says nothing about later times, so (1) is false.
2. Velocity can be zero while acceleration is zero (rest): (2) true.
3. Zero acceleration over an interval means constant velocity: (3) true. Zero speed over an interval means velocity is constant (zero), so acceleration is zero: (4) true.

Teacher's Note:
1. An instant value of acceleration does not fix the velocity, but acceleration over an interval does fix how velocity changes in it.

Answer: (c) (1), (2) and (4) only

Explanation:
1. Opposite signs of \( v,a \) mean the speed decreases (1 true). Opposite signs of \( x,v \) mean the position magnitude shrinks (2 true).
2. At the top of a thrown ball \( v=0 \) but \( a=-g \), so (3) is false.
3. If \( v=0 \) throughout an interval, its derivative is zero throughout it, so (4) is true.

Teacher's Note:
1. The turning point of a thrown ball is the classic example of zero velocity with non-zero acceleration.

Answer: (a) (3) and (4) only

Explanation:
1. \( v=u+2a(t-2) \), so at \( t=0 \) it is \( u-4a \) (1 false).
2. Acceleration \( =2a \) (3 true, 2 false); at \( t=2 \): \( y=0 \) (4 true).

Teacher's Note:
1. Differentiate the given expression and substitute carefully; the shift in time origin is the trap.

Answer: (d) \( -10\ \text{m/s} \)

Explanation:
1. \( v_{av}=\frac{\Delta x}{\Delta t}=\frac{-100-100}{20}=-10\ \text{m/s} \).

Teacher's Note:
1. Average velocity depends only on end positions and the time taken, not on the shape of the curve.

Answer: (c) zero

Explanation:
1. The slope of the tangent gives the instantaneous velocity. Equal angles at the start and at the end mean equal initial and final velocities.
2. So \( \Delta v=0 \) and the average acceleration is zero.

Teacher's Note:
1. Average acceleration depends only on the change in velocity between the end points.

Answer: (b) \( \dfrac{2n-1}{2n+1} \)

Explanation:
1. Distance in the nth second from rest: \( S_n=\frac a2(2n-1) \).
2. So \( \frac{S_n}{S_{n+1}}=\frac{2n-1}{2n+1} \).

Teacher's Note:
1. The distances in successive seconds from rest are in the ratio 1:3:5:7 and so on.

Answer: (a) \( 55\ \text{m/s} \)

Explanation:
1. The acceleration is positive throughout, so the speed keeps rising until \( t=11 \) s.
2. Speed = area of the triangle \( =\tfrac12\times11\times10=55\ \text{m/s} \).

Teacher's Note:
1. The area under the acceleration-time graph is the change in velocity.
2. The maximum speed comes where acceleration ends.

Answer: (d) 2 s

Explanation:
1. \( \frac{dv}{\sqrt v}=-2.5\,dt \).
2. Integrate: \( 2\sqrt v\Big|_{6.25}^{0}=-2.5t \), i.e. \( 2(2.5)=2.5t \).
3. So \( t=2 \) s.

Teacher's Note:
1. Separate the variables v and t, then integrate between the initial and final speeds.

Answer: (c) \( 2gH=nu^2(n-2) \)

Explanation:
1. Time to the top \( u/g \), total time \( nu/g \).
2. Displacement is \( -H \): \( -H=u\frac{nu}{g}-\frac12g\frac{n^2u^2}{g^2}=\frac{nu^2}{g}-\frac{n^2u^2}{2g} \).
3. So \( 2gH=nu^2(n-2) \).

Teacher's Note:
1. Substitute the given total time into the displacement equation and rearrange.

Answer: (b) A single straight line of negative slope, starting at a positive value and crossing the time axis

Explanation:
1. \( v=u-gt \) is a straight line with slope \( -g \) throughout the flight.
2. It starts positive, is zero at the top and becomes negative while the body falls, with no change of slope.

Teacher's Note:
1. The acceleration is constant, so the v-t graph is a single straight line of slope minus g.

Answer: (a) Distance travelled versus time: an S-shaped curve that levels off to a constant value

Explanation:
1. Distance travelled never decreases and keeps increasing as long as the ball is moving, so it cannot level off while the ball is still in motion.
2. The other three graphs are correct: \( x=ut-gt^2/2 \) is a parabola, \( v=u-gt \) is a line, and \( v=\pm\sqrt{u^2-2gx} \) has two branches.

Teacher's Note:
1. Distance is the accumulated path length, always rising in time.
2. Position and velocity can fall and change sign.

Answer: (d) \( \dfrac{b^2\tau}{2} \)

Explanation:
1. \( \frac{dx}{dt}=b\sqrt x \) gives \( 2\sqrt x=bt \), so \( x=\frac{b^2t^2}{4} \).
2. Speed \( =b\sqrt x=\frac{b^2t}{2} \), at \( t=\tau \): \( \frac{b^2\tau}{2} \).

Teacher's Note:
1. Integrate dx over square root of x to relate x and t, then substitute back into the speed.

Answer: (c) 9 m

Explanation:
1. 0 to 2 s: \( \tfrac12\times2\times2=2 \) m; 2 to 4 s: \( 2\times2=4 \) m; 4 to 5 s: \( 3\times1=3 \) m.
2. Position \( 2+4+3=9 \) m.

Teacher's Note:
1. Position equals the area under the velocity-time graph from the start.

Answer: (b) 3

Explanation:
1. Differentiate: \( x\dot x=at+b \) and again \( \dot x^2+x\ddot x=a \).
2. So \( \ddot x=\frac{a-\dot x^2}{x}=\frac{ax^2-(at+b)^2}{x^3}=\frac{ac-b^2}{x^3} \), i.e. \( x^{-3} \).

Teacher's Note:
1. The numerator turns out constant because (at+b) squared is a x squared minus (ac minus b squared).

Answer: (a) 60 m

Explanation:
1. Y motion: \( 32=\tfrac12(4)t^2 \), so \( t=4 \) s.
2. X motion: \( x=3(4)+\tfrac12(6)(16)=12+48=60 \) m.

Teacher's Note:
1. The x and y motions are independent and share only the time.

Answer: (d) (q) and (s) only

Explanation:
1. \( v=kx \) gives \( a=v\frac{dv}{dx}=k^2x \).
2. So the acceleration is proportional to \( x \) and is therefore non-uniform: (q) and (s).

Teacher's Note:
1. Use a = v dv/dx when velocity is given as a function of position.

Answer: (c) (p) only

Explanation:
1. \( v^2=kx \) gives \( 2v\frac{dv}{dx}=k \), so \( a=v\frac{dv}{dx}=\frac k2 \).
2. The acceleration is constant: (p).

Teacher's Note:
1. The relation v squared = 2 a x for constant acceleration from rest is a straight line in v squared versus x.

Answer: (b) (p) only

Explanation:
1. \( v=kt \) gives \( a=dv/dt=k \), a constant.
2. So only (p) is correct.

Teacher's Note:
1. The slope of a v-t graph is the acceleration; a straight line has constant slope.

Answer: (a) (q) and (r) only

Explanation:
1. \( v=kt^2 \) gives \( a=2kt \).
2. The acceleration is proportional to \( t \), hence non-uniform: (q) and (r). Since \( x=kt^3/3 \), it is not proportional to \( x \).

Teacher's Note:
1. Differentiate the given relation with respect to time to read off the acceleration law.

Answer: (a) \( \dfrac{\ell}{\sqrt{\eta^2-1}} \)

Explanation:
1. Let the turn-off point C be at distance \( d \) from D and the highway speed be \( v \) (field speed \( v/\eta \)). Time \( t=\frac{AD-d}{v}+\frac{\eta\sqrt{d^2+\ell^2}}{v} \).
2. Minimise: \( \frac{dt}{dd}=-1+\frac{\eta d}{\sqrt{d^2+\ell^2}}=0 \).
3. So \( d^2(\eta^2-1)=\ell^2 \) and \( d=\frac{\ell}{\sqrt{\eta^2-1}} \).

Teacher's Note:
1. Set the derivative of the total time to zero; this is the same condition as Snell's law for the fastest path.