JEE Physics Motion In A Straight Line MCQs Set 21

Answer: (b) \( \frac{x_1}{x_2}=\frac{\beta}{\alpha}=\frac{t_1}{t_2} \)

Explanation:
1. The maximum speed \( v \) is reached at the end of the first stage: \( v=\alpha t_1=\beta t_2 \), so \( \frac{t_1}{t_2}=\frac{\beta}{\alpha} \).
2. Using \( v^2=2\alpha x_1=2\beta x_2 \) gives \( \frac{x_1}{x_2}=\frac{\beta}{\alpha} \).
3. Therefore \( \frac{x_1}{x_2}=\frac{\beta}{\alpha}=\frac{t_1}{t_2} \).

Teacher's Note:
1. The same peak speed links the two stages. The larger the acceleration, the shorter the time and distance needed to reach that speed.

Answer: (b) \( v_1-v_2 : v_2-v_3=t_1+t_2 : t_2+t_3 \)

Explanation:
1. For uniform acceleration the average velocity over an interval equals the instantaneous velocity at its mid-instant.
2. The mid-instants of the intervals are separated by \( \frac{t_1+t_2}{2} \) and \( \frac{t_2+t_3}{2} \).
3. The velocity changes by \( a\times \) (time gap), so \( v_1-v_2 : v_2-v_3=(t_1+t_2) : (t_2+t_3) \).

Teacher's Note:
1. Average velocity over an interval is the velocity at the middle of that interval whenever acceleration is constant. Then differences of velocities are proportional to the gaps between mid-instants.

Answer: (c) in the ratio of the differences of square roots of integers \( (\sqrt{1}-\sqrt{0}):(\sqrt{2}-\sqrt{1}):(\sqrt{3}-\sqrt{2}):\dots \)

Explanation:
1. Time to fall \( n \) metres from rest: \( t_n=\sqrt{\frac{2n}{g}} \).
2. Time for the \( n \)th metre alone: \( t_n-t_{n-1}=\sqrt{\frac{2}{g}}\left(\sqrt{n}-\sqrt{n-1}\right) \).
3. So the times are in the ratio \( (\sqrt1-\sqrt0):(\sqrt2-\sqrt1):(\sqrt3-\sqrt2)\dots \).

Teacher's Note:
1. A falling body speeds up, so each successive metre takes less time. The total times go as square roots, and the individual times as differences of square roots.

Answer: (a) \( \frac{1}{2}gt_1t_2 \)

Explanation:
1. At height \( h \): \( h=ut-\frac{1}{2}gt^2 \Rightarrow \frac{1}{2}gt^2-ut+h=0 \).
2. The two roots are \( t_1 \) and \( t_2 \), so their product is \( t_1t_2=\frac{2h}{g} \).
3. Therefore \( h=\frac{1}{2}gt_1t_2 \).

Teacher's Note:
1. The quadratic in time has two roots, the up-time and the down-time. Product of roots gives the result directly.

Answer: (d) \( v=u+\frac{1}{2}at^2 \)

Explanation:
1. The acceleration is not constant: \( \frac{dv}{dt}=at \).
2. Integrating from 0 to \( t \): \( v-u=\int_0^t at\,dt=\frac{1}{2}at^2 \).
3. So \( v=u+\frac{1}{2}at^2 \).

Teacher's Note:
1. The formula \( v=u+at \) is only for constant acceleration. When acceleration depends on time, integrate it.

Answer: (d) 4 cm/s²

Explanation:
1. Acceleration is the slope of the graph in each segment.
2. 0 to 20 s: \( 20/20=1 \). 20 to 30 s: 0. 30 to 40 s: \( \frac{60-20}{10}=4 \). 40 to 70 s: \( -\frac{60}{30}=-2 \).
3. The maximum acceleration is 4 cm/s².

Teacher's Note:
1. Read slopes segment by segment. The steepest straight part of a velocity-time graph gives the greatest acceleration magnitude.

Answer: (d) \( \frac{v_0^2}{2m} \)

Explanation:
1. Distance = area under the graph = \( \frac{1}{2}v_0T \).
2. Slope \( m=\frac{v_0}{T} \Rightarrow T=\frac{v_0}{m} \).
3. Distance = \( \frac{1}{2}v_0\cdot\frac{v_0}{m}=\frac{v_0^2}{2m} \).

Teacher's Note:
1. The area under a velocity-time graph is the distance (when velocity does not change sign). Express the answer using only the quantities asked for.

Answer: (b) 15 m

Explanation:
1. The motion is symmetric about the instant of maximum height, \( t=\frac{1+6}{2}=3.5 \) s.
2. At \( t=1 \) and 6 s the ball is \( 2.5 \) s from the top; at \( t=2 \) and 5 s it is \( 1.5 \) s from the top.
3. \( h=\frac{1}{2}(7.5)\left(2.5^2-1.5^2\right)=\frac{1}{2}(7.5)(4)=15 \) m.

Teacher's Note:
1. Measure everything from the highest point, where velocity is zero. The depth below the top is \( \frac12gt^2 \) for time \( t \) measured from the peak.

Answer: (b) cover a distance of 140 m in reaching the ground

Explanation:
1. At release: height \( =\frac12(2)(10)^2=100 \) m and upward speed \( v=2\times10=20 \) m/s.
2. The stone goes further up by \( \frac{20^2}{2\times10}=20 \) m, then falls \( 120 \) m: distance \( =20+120=140 \) m.
3. Its displacement is 100 m downward, and the time to land is \( 2+\sqrt{24}\approx6.9 \) s. It first keeps moving up.

Teacher's Note:
1. A released object keeps the velocity it had at the moment of release. The balloon's speed is therefore the stone's initial upward speed.

Answer: (c) 15 m/s, 5 m/s

Explanation:
1. The trains must cover a combined length of 200 m in each case.
2. Overtaking (same direction): \( v_A-v_B=\frac{200}{20}=10 \). Crossing (opposite direction): \( v_A+v_B=\frac{200}{10}=20 \).
3. Solving, \( v_A=15 \) m/s and \( v_B=5 \) m/s.

Teacher's Note:
1. Use relative velocity: subtract speeds for the same direction and add them for opposite directions. The distance to be covered is the sum of the lengths.

Answer: (a) \( u \)

Explanation:
1. Velocity of A at time \( t \): \( u-gt \) (upward). Velocity of B: \( -gt \) (upward).
2. Relative velocity = \( (u-gt)-(-gt)=u \).
3. The relative speed is constant and equals \( u \).

Teacher's Note:
1. Both balls have the same acceleration, so in the frame of one the other moves with constant velocity. This idea simplifies many meeting-time problems.

Answer: (b) \( s=\frac12ft^2 \)

Explanation:
1. Let the top speed be \( v \): \( v^2=2fs \). Distance while decelerating \( =\frac{v^2}{2(f/2)}=\frac{v^2}{f}=2s \).
2. Total: \( s+vt+2s=5s \Rightarrow vt=2s \).
3. Then \( v^2t^2=4s^2 \Rightarrow 2fs\,t^2=4s^2 \Rightarrow s=\frac12ft^2 \).

Teacher's Note:
1. Break the journey into three stages and note the top speed is common. A velocity-time trapezium helps visualise the total distance.

Answer: (b) 2 m/s, 6 m/s

Explanation:
1. First second: \( a=12 \) m/s², displacement \( =6 \) m. After reversal \( a=-12 \) m/s²: the car slows to rest in 1 s (displacement \( +6 \) m) and then speeds up backward in the next second (displacement \( -6 \) m).
2. Net displacement \( =6+6-6=6 \) m, so average velocity \( =\frac63=2 \) m/s.
3. Total distance \( =6+6+6=18 \) m, so average speed \( =\frac{18}{3}=6 \) m/s.

Teacher's Note:
1. Split the motion where the velocity changes sign. Displacement can partly cancel, but distance always adds up.

Answer: (c) 4 s

Explanation:
1. Take upward as positive: \( u=+10 \) m/s, \( a=-10 \) m/s², and final displacement \( s=-40 \) m.
2. \( -40=10t-5t^2 \Rightarrow t^2-2t-8=0 \).
3. \( (t-4)(t+2)=0 \), so the positive root is \( t=4 \) s.

Teacher's Note:
1. Sign convention matters: the ground is below the starting point, so the displacement is negative. Reject the negative time root.

Answer: (a) 5 cm/s and 37 cm/s

Explanation:
1. \( v=\frac{ds}{dt}=8t+5 \).
2. At \( t=0 \): \( v=5 \) cm/s.
3. At \( t=4 \) s: \( v=8(4)+5=37 \) cm/s.

Teacher's Note:
1. Velocity is the first derivative of displacement. Initial speed means the value at \( t=0 \).

Answer: (a) \( \frac{8H}{T_P^2-T_Q^2} \)

Explanation:
1. If the highest point is \( h \) above Q, then \( T_Q=2\sqrt{\frac{2h}{g}} \) and, since P is lower by \( H \), \( T_P=2\sqrt{\frac{2(h+H)}{g}} \).
2. Squaring: \( T_Q^2=\frac{8h}{g} \) and \( T_P^2=\frac{8(h+H)}{g} \).
3. Subtracting, \( T_P^2-T_Q^2=\frac{8H}{g} \), hence \( g=\frac{8H}{T_P^2-T_Q^2} \).

Teacher's Note:
1. Time between two passes of a level equals twice the free-fall time from the top to that level. Subtracting the squares removes the unknown height.