JEE Physics Motion In A Straight Line MCQs Set 06

Answer: (b) \( \frac12gt^2 \)

Explanation:
1. Reverse the motion: the last t seconds of the ascent are the first t seconds of a fall from rest at the top.
2. Distance \( =\frac12gt^2 \).

Teacher's Note:
1. Time reversal turns the ascent into a free fall.

Answer: (a) 29 s

Explanation:
1. In every 8 steps (8 s) he advances 2 m.
2. After 3 such cycles (24 s) he is at 6 m. The next 5 forward steps take him to 11 m, at 29 s, where he falls into the pit.
3. He does not reach 11 m earlier because the maximum position in each cycle is 2n + 5 m, which equals 11 m only for n = 3.

Teacher's Note:
1. Find the net gain per cycle and the maximum position within a cycle.

Answer: (c) \( \frac32A+\frac73B \)

Explanation:
1. \( s=\int_1^2(At+Bt^2)dt=\frac A2(4-1)+\frac B3(8-1) \).
2. \( =\frac32A+\frac73B \).

Teacher's Note:
1. Integrate the velocity between the limits.

Answer: (b) 49 m/s

Explanation:
1. The distances in the 5th and 6th seconds are equal when the highest point is reached exactly at the end of the 5th second (the body rises for 1 s and falls for 1 s over this period).
2. \( u=g\times5=9.8\times5=49 \) m/s.

Teacher's Note:
1. The two seconds are symmetric about the top.

Answer: (c) \( 7\sqrt2 \) units

Explanation:
1. \( \vec v=(3+4)\hat i+(4+3)\hat j=7\hat i+7\hat j \).
2. Speed \( =7\sqrt2 \).

Teacher's Note:
1. Use v = u + at in vector form.

Answer: (a) a horizontal straight line

Explanation:
1. Uniform motion means constant velocity, which is a horizontal line in the v-t graph.
2. An inclined line represents uniform acceleration.

Teacher's Note:
1. Uniform means that the velocity does not change.

Answer: (c) 32 m/s\( ^2 \)

Explanation:
1. \( a=\frac{d^2s}{dt^2}=18t+14 \).
2. At \( t=1 \) s: \( a=32 \) m/s\( ^2 \).

Teacher's Note:
1. Differentiate twice.

Answer: (c) 500 m/s

Explanation:
1. Mean velocity \( =\frac{60000}{120}=500 \) m/s.

Teacher's Note:
1. Divide the displacement by the time.

Answer: (a) \( \frac{10}3 \) m

Explanation:
1. \( s_n=\frac a2(2n-1)=\frac23\times5=\frac{10}3 \) m.

Teacher's Note:
1. Use the nth second formula.

Answer: (b) \( T=\sqrt{\frac{2h}g}+\frac hv \)

Explanation:
1. Fall time \( t_1=\sqrt{\frac{2h}g} \).
2. Sound takes \( t_2=\frac hv \) to come up.
3. \( T=t_1+t_2 \).

Teacher's Note:
1. Add the fall time and the time for sound to travel up.

Answer: (b) 18 cm/s

Explanation:
1. \( v=2bt=2\times3\times3=18 \) cm/s.

Teacher's Note:
1. The constant a does not affect the velocity.

Answer: (d) straight line

Explanation:
1. \( v=u-gt \) is a linear function of t with constant slope \( -g \).
2. So the graph is a straight line.

Teacher's Note:
1. Constant acceleration gives a straight line v-t graph.

Answer: (b) 15 m

Explanation:
1. The motion is symmetric about \( t=3.5 \) s. Heights below the top are \( \frac12g(\Delta t)^2 \).
2. At 1 s and 6 s: \( \Delta t=2.5 \); at 2 s and 5 s: \( \Delta t=1.5 \).
3. \( h=\frac12(7.5)(2.5^2-1.5^2)=3.75\times4=15 \) m.

Teacher's Note:
1. Use the symmetry of the up and down journey.

Answer: (d) -3084

Explanation:
1. Using \( v^2=u^2+2as \): \( a=\frac{v^2-u^2}{2s}=\frac{900-4600}{2\times0.6}=-3083.3\approx-3084 \) km/h\( ^2 \).
2. The negative sign shows retardation.

Teacher's Note:
1. Read the two points from the graph and use the third equation of motion.

Answer: (b) 20 m/s

Explanation:
1. \( v=u-gt=40-20=20 \) m/s (upwards).

Teacher's Note:
1. Use v = u - g t.

Answer: (c) 24.5 m

Explanation:
1. The first body has dropped for 3 s: \( 4.9\times9=44.1 \) m.
2. The second has dropped for 2 s: \( 4.9\times4=19.6 \) m.
3. Separation \( =24.5 \) m.

Teacher's Note:
1. Find each distance covered separately.

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

Explanation:
1. \( v=u+\int_0^tat\,dt=u+\frac{at^2}2 \).

Teacher's Note:
1. Integrate the acceleration.

Answer: (b) A body can have a constant velocity and still have a varying speed

Explanation:
1. At the top of a vertical throw the velocity is zero but the acceleration is g, so (a) is true.
2. Uniform circular motion has constant speed and changing velocity, so (c) is true; projectile motion changes direction with constant g, so (d) is true.
3. Speed is the magnitude of velocity, so a constant velocity means a constant speed: (b) is false.

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

Answer: (b) \( \frac{2v_dv_u}{v_d+v_u} \)

Explanation:
1. Total distance \( 2d \); total time \( \frac d{v_u}+\frac d{v_d} \).
2. Average speed \( =\frac{2v_uv_d}{v_u+v_d} \).

Teacher's Note:
1. Use total distance over total time.

Answer: (b) \( \sqrt{2gd} \)

Explanation:
1. Starting from rest, \( v^2=2gd \).
2. The change in speed is \( \sqrt{2gd} \).

Teacher's Note:
1. The initial speed is zero at the highest point.

Answer: (c) At \( \frac{3H}4 \) metres from the ground

Explanation:
1. Distance dropped in time \( \frac t2 \): \( \frac12g\frac{t^2}4=\frac14\left(\frac12gt^2\right)=\frac H4 \).
2. Height above the ground \( =H-\frac H4=\frac{3H}4 \).

Teacher's Note:
1. Distance dropped is proportional to the square of time.

Answer: (b) 78.4 m

Explanation:
1. \( h_1=\frac12(9.8)(5)^2=122.5 \) m and \( h_2=\frac12(9.8)(3)^2=44.1 \) m.
2. Difference \( =78.4 \) m.

Teacher's Note:
1. The second ball falls for only 3 s.

Answer: (b) Area A + Area C - Area B - Area D

Explanation:
1. Areas above the time axis give positive displacement and areas below give negative displacement.
2. Displacement \( =A+C-B-D \).

Teacher's Note:
1. Regions below the axis count negatively.