JEE Physics Motion In A Straight Line MCQs Set 23

Answer: (b) 48 m

Explanation:
1. Semicircle of radius \( r \): \( \pi r=11 \) m. Semicircle of radius \( 2r \): \( 2\pi r=22 \) m.
2. Distance \( =7+11+8+22=48 \) m.

Teacher's Note:
1. Distance counts the whole length of the path, including curved parts.
2. Displacement counts only the straight line from start to end.

Answer: (a) 36 m

Explanation:
1. The diameters and straight sections lie end to end along the line PR.
2. Displacement \( =7+2r+8+4r=15+6r=15+21=36 \) m.

Teacher's Note:
1. Displacement is the straight-line separation of the end points, here the sum of all the collinear pieces: 7, 2r, 8 and 4r.

Answer: (d) 110 m and 50 m

Explanation:
1. Distance \( =50+40+20=110 \) m.
2. Net: 30 m south and 40 m west, so displacement \( =\sqrt{30^2+40^2}=50 \) m, at \( \tan^{-1}\frac43 \) west of south.

Teacher's Note:
1. Add up all lengths for distance, but combine the north-south parts as signed quantities for displacement.

Answer: (c) 32 km/h and zero

Explanation:
1. Distance \( =12416-12352=64 \) km in 2 h: average speed \( =32 \) km/h.
2. He ends where he began, so displacement and average velocity are zero.

Teacher's Note:
1. The odometer measures distance, not displacement.
2. The 15-minute stay is included in the total time.

Answer: (b) \( \dfrac{3v_1v_2v_3}{v_1v_2+v_2v_3+v_3v_1} \)

Explanation:
1. Let the total distance be \( 3s \). Times: \( \frac s{v_1},\frac s{v_2},\frac s{v_3} \).
2. Average speed \( =\frac{3s}{s(1/v_1+1/v_2+1/v_3)}=\frac{3v_1v_2v_3}{v_1v_2+v_2v_3+v_3v_1} \) (harmonic mean).

Teacher's Note:
1. Equal distances give a harmonic mean of the speeds, while equal times give the arithmetic mean.
2. Always start from total distance over total time.

Answer: (a) \( A+300B \)

Explanation:
1. \( v=\frac{dx}{dt}=A+12Bt^2 \).
2. At \( t=5 \): \( v=A+12B(25)=A+300B \).

Teacher's Note:
1. Differentiate position once for velocity and again for acceleration (here 24Bt, which is 120B at 5 s).

Answer: (d) \( 2.5\ \text{m/s}^2 \)

Explanation:
1. \( 18\ \text{km/h}=5\ \text{m/s} \).
2. Average acceleration \( =\frac{5-0}{2}=2.5\ \text{m/s}^2 \).

Teacher's Note:
1. Convert km/h to m/s by dividing by 3.6 before using any formula.

Answer: (c) 50 m

Explanation:
1. East: \( 5\times2=10 \) m. Then south \( 40 \) m. Then west \( 5\times8=40 \) m.
2. Net: \( 30 \) m west and \( 40 \) m south, so displacement \( =\sqrt{30^2+40^2}=50 \) m.

Teacher's Note:
1. Right turns from east take the boy south and then west.
2. Treat the legs as perpendicular vectors.

Answer: (b) \( \frac{25}{9}\ \text{m/s} \)

Explanation:
1. Total time \( 2+8+8=18 \) s and displacement \( 50 \) m (total distance \( 90 \) m).
2. Average velocity \( =\frac{50}{18}=\frac{25}{9}\ \text{m/s} \), while the average speed is \( 90/18=5 \) m/s.

Teacher's Note:
1. Average velocity uses displacement and average speed uses distance.
2. They differ whenever the path turns.

Answer: (a) 87.5 m

Explanation:
1. Speeds \( 10\ \text{m/s} \) and \( 25\ \text{m/s} \): \( a=\frac{15}{5}=3\ \text{m/s}^2 \).
2. Distance \( =\frac{10+25}{2}\times5=87.5 \) m.

Teacher's Note:
1. For constant acceleration the average velocity is the mean of initial and final velocities.

Answer: (d) 2.7 km

Explanation:
1. Acceleration phase: \( v=2(30)=60\ \text{m/s} \), distance \( \tfrac12(2)(30)^2=900 \) m.
2. Braking from 60 m/s to rest in 60 s: distance \( \frac{60}{2}\times60=1800 \) m.
3. Total \( 900+1800=2700 \) m.

Teacher's Note:
1. The area of the triangle in the speed-time graph is the total distance: half the base 90 s times the height 60 m/s.

Answer: (c) 225 m and 2.25 km

Explanation:
1. Maximum speed \( 60\ \text{m/s} \), half is \( 30 \) m/s. While accelerating: \( x=\frac{30^2}{2\times2}=225 \) m.
2. Braking retardation \( 1\ \text{m/s}^2 \) starting at 900 m: \( x=900+\frac{60^2-30^2}{2\times1}=2250 \) m.
3. So 225 m and 2.25 km.

Teacher's Note:
1. Half the maximum speed occurs once on the way up and once on the way down, so two positions are needed.

Answer: (b) 15 m

Explanation:
1. \( u=20\ \text{m/s} \). Use \( s_n=u-\frac a2(2n-1) \) with \( n=3 \).
2. So \( s_3=20-\frac22(5)=15 \) m. (It stops after 100 m in 10 s.)

Teacher's Note:
1. Displacement in the nth second is the position at n seconds minus the position at n minus 1 seconds, which gives the formula u plus a over 2 times (2n minus 1).

Answer: (a) 20 m

Explanation:
1. \( s_n=\frac g2(2n-1)=5(2n-1)=15 \) so \( n=2 \) s.
2. Height \( =\tfrac12(10)(2)^2=20 \) m.

Teacher's Note:
1. The distance in the last second fixes the total time.
2. Then use h = half g t squared.

Answer: (d) \( 20\ \text{m/s} \)

Explanation:
1. The ball rises 20 m and stops: \( u^2=2gh=400 \).
2. So \( u=20\ \text{m/s} \) (reached after \( 2 \) s).

Teacher's Note:
1. Coming to rest at a given height means the final velocity is zero there, so v squared = u squared minus 2 g h applies directly.

Answer: (c) 4 m

Explanation:
1. The ball reaches 20 m with zero speed at \( t=\frac{20}{10}=2 \) s.
2. The plane then has moved \( s=\tfrac12(2)(2)^2=4 \) m.

Teacher's Note:
1. Both objects share the same elapsed time, so the ball's time of rise gives the plane's displacement.

Answer: (b) \( \sqrt{656}\ \text{m} \)

Explanation:
1. Plane speed at \( t=2 \) s is \( 4 \) m/s. In the next 2 s: \( 4(2)+\tfrac12(4)(2)^2=16 \) m, so it is \( 16 \) m horizontally from the man.
2. The ball returns after another \( 2 \) s. Distance \( =\sqrt{16^2+20^2}=\sqrt{656} \) m.

Teacher's Note:
1. Horizontal and vertical separations are perpendicular, so combine them with Pythagoras.

Answer: (a) 10 m and 6 m

Explanation:
1. Area from 0 to 2 s: \( \tfrac12\times2\times4=4 \) m; from 2 to 3 s: \( 4 \) m; from 3 to 5 s (below the axis): \( \tfrac12\times2\times2=2 \) m.
2. Distance \( 4+4+2=10 \) m; displacement \( 4+4-2=6 \) m.

Teacher's Note:
1. Areas below the time axis count as negative for displacement but positive for distance.

Answer: (d) 2000 m

Explanation:
1. Speed after 10 s is \( 100 \) m/s: distance \( \tfrac12(10)(10)^2=500 \) m.
2. From 10 s to 20 s at constant \( 100 \) m/s: \( 1000 \) m.
3. Last 10 s decelerates from 100 to 0: \( 500 \) m. Total \( 2000 \) m.

Teacher's Note:
1. Build the speed-time graph from the acceleration steps: a trapezoid with area 2000 m.

Answer: (c) B covers the larger distance

Explanation:
1. Total time \( t \). For A: \( s_A=\tfrac12a(\tfrac t2)^2+(a\tfrac t2)\tfrac t2+\tfrac12(2a)(\tfrac t2)^2=\tfrac58at^2 \).
2. For B: \( s_B=\tfrac12(2a)(\tfrac t2)^2+(at)\tfrac t2+\tfrac12a(\tfrac t2)^2=\tfrac78at^2 \).
3. So B covers more distance.

Teacher's Note:
1. Whoever builds up the speed earlier covers more ground in the second half.
2. A speed-time sketch makes this obvious.

Answer: (b) \( 3\sqrt5\ \text{s} \)

Explanation:
1. Total time \( T \): the speed after \( T/3 \) is \( 8T/3 \).
2. Distance \( \tfrac12(8)(\tfrac T3)^2+\tfrac{8T}{3}\cdot\tfrac{2T}{3}=\tfrac{20T^2}{9}=100 \).
3. So \( T^2=45 \) and \( T=3\sqrt5\ \text{s} \).

Teacher's Note:
1. Split the race into the accelerating and uniform parts, then express both distances in terms of the single unknown total time.

Answer: (a) \( \dfrac{2v(n-1)}{n} \)

Explanation:
1. \( a=v/n \). Velocity two seconds before the end is \( v-2a=v-\frac{2v}{n} \).
2. Displacement \( =\frac{v+(v-2v/n)}{2}\times2=\frac{2v(n-1)}{n} \).

Teacher's Note:
1. Use average velocity times time for any interval of uniform acceleration.

Answer: (d) 1 : 3 : 5

Explanation:
1. From rest, distances in successive equal intervals are in the ratio \( 1:3:5 \) (Galileo's odd-number rule).
2. Check: \( \tfrac12a(10)^2=50a,\ \tfrac12a(400-100)=150a,\ \tfrac12a(900-400)=250a \).

Teacher's Note:
1. Positions at 10, 20, 30 s are in the ratio 1:4:9, so the differences are 1:3:5.

Answer: (c) \( v_T+v_B=12\ \text{m/s} \)

Explanation:
1. Average speed across the window \( =\frac{3}{0.5}=6 \) m/s, and it equals \( \frac{v_T+v_B}{2} \), so \( v_T+v_B=12 \) m/s.
2. Also \( v_B-v_T=g(0.5)=4.9 \) m/s, giving \( v_T=3.55,\ v_B=8.45 \) m/s.

Teacher's Note:
1. In uniform acceleration, average speed equals half the sum of end speeds.
2. Use it with the time of passage.

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

Explanation:
1. Time for the stone to fall: \( t_1=\sqrt{2h/g} \).
2. Time for the sound to rise: \( t_2=h/v \).
3. Total \( T=t_1+t_2 \).

Teacher's Note:
1. Two consecutive stages, each with its own speed law, add up in time.

Answer: (a) \( \frac{275}{3}\ \text{m/s} \)

Explanation:
1. Student's fall time \( \sqrt{2(320)/10}=8 \) s, so superman has \( 3 \) s.
2. Superman: \( 320=3u+\tfrac12(10)(3)^2=3u+45 \).
3. So \( u=\frac{275}{3}\ \text{m/s} \).

Teacher's Note:
1. Both bodies cover the same height, but superman has less time.
2. Solve for the initial speed that makes up the 5-second delay.

Answer: (d) 125 m

Explanation:
1. The student cannot be saved if he reaches the ground before superman leaves the roof, i.e. within 5 s.
2. Height \( =\tfrac12(10)(5)^2=125 \) m or less.

Teacher's Note:
1. Superman can only help if the student is still falling when he starts, so the limiting case is a fall time of exactly 5 s.

Answer: (c) \( 10.5\ \text{m/s} \)

Explanation:
1. The acceleration is \( a=5-t \).
2. Speed = area under the a-t graph \( =\int_0^3(5-t)\,dt=15-4.5=10.5\ \text{m/s} \).

Teacher's Note:
1. The area under an acceleration-time graph gives the change in velocity.
2. Integrate the straight-line function between the limits.

Answer: (b) \( \dfrac13 \)

Explanation:
1. Velocity is the slope: \( v_A=\tan30^\circ=\frac1{\sqrt3} \) and \( v_B=\tan60^\circ=\sqrt3 \).
2. Ratio \( =\frac{1/\sqrt3}{\sqrt3}=\frac13 \).

Teacher's Note:
1. Slope of a displacement-time line is tan of the angle with the time axis, not sin or cos.