JEE Physics Motion In A Straight Line MCQs Set 15

Answer: (a)

Explanation:
1. From 0 to 2 s, \(a=3\) m/s², so v rises linearly from 0 to 6 m/s
2. From 2 s to 4 s, \(a=-3\) m/s², so v falls linearly from 6 m/s back to 0
3. Only graph (a) shows this continuous triangular variation

Teacher's Note:
1. Constant acceleration gives a straight-line v-t graph with that slope.
2. Velocity cannot jump suddenly when the acceleration is finite.

Answer: (a)

Explanation:
1. 0 to 2 s: \(a=5\), v rises from 0 to 10 m/s
2. 2 s to 6 s: \(a=0\), v stays 10 m/s
3. 6 s to 10 s: \(a=-5\), v falls by 20 m/s to -10 m/s
4. This matches graph (a)

Teacher's Note:
1. Zero acceleration gives a horizontal v-t line.
2. Start from the initial velocity and add the area of each a-t block.

Answer: (d) 60 km

Explanation:
1. The velocity becomes zero at t = 120 s, so the maximum height is reached then
2. Height \(=\) area of the triangle from 0 to 120 s \(=\tfrac12(120)(1000)\)
3. \(h=60000\) m \(=60\) km

Teacher's Note:
1. The highest point is where the velocity is zero, not where it is largest.
2. Convert metres to kilometres at the end.

Answer: (c) 500 m/s

Explanation:
1. Displacement up to the highest point \(=60\) km \(=60000\) m, covered in 120 s
2. Mean velocity \(=\dfrac{60000}{120}=500\) m/s

Teacher's Note:
1. Mean velocity is total displacement divided by total time.
2. The motion is one-directional until the highest point, so this also equals the average speed.

Answer: (d) 10 m/s²

Explanation:
1. After burn-out the velocity falls from 1000 m/s at 20 s to 0 at 120 s
2. Retardation \(=\dfrac{1000-0}{120-20}=10\) m/s², which equals g

Teacher's Note:
1. Retardation is the magnitude of negative acceleration.
2. The slope of line AB equals g, so only gravity acts after burn-out.

Answer: (d) OAB and BC

Explanation:
1. The velocity is positive from O to B, so the rocket keeps moving up along OA and AB
2. After B (t = 120 s) the velocity is negative, so it comes down along BC

Teacher's Note:
1. The sign of velocity tells the direction of motion.
2. A (the point of maximum speed) is not the highest point; B is.

Answer: (d)

Explanation:
1. Uniform acceleration from rest gives \(x=\tfrac12at^2\), a parabola whose slope increases from zero
2. Uniform velocity afterwards gives a straight line with the slope reached at 4 s, which is steeper than the beginning
3. Only graph (d) shows this

Teacher's Note:
1. The slope of an x-t graph is velocity.
2. Constant velocity means a straight line; increasing velocity means a curve getting steeper.

Answer: (d) -3084

Explanation:
1. \(v^2=u^2+2as\), so the slope of the v²-s graph is \(2a\)
2. Slope \(=\dfrac{900-4600}{0.6}\approx-6167\) (km/h)²/km
3. \(a=\dfrac{-6167}{2}\approx-3083\) km/h², closest to -3084

Teacher's Note:
1. On a v²-s graph the slope equals twice the acceleration.
2. The negative slope means the particle is retarding.

Answer: (d) 10.7 m/s

Explanation:
1. Total distance \(=\tfrac12v(40)+v(180)+\tfrac12v(20)\) (accelerating, uniform, retarding)
2. \(20v+180v+10v=210v=2250\) m
3. \(v=\dfrac{2250}{210}\approx10.7\) m/s

Teacher's Note:
1. Draw the trapezium v-t graph: total area equals the distance.
2. Total time is 240 s, so the constant-speed part lasts 180 s.

Answer: (d) 145 km/h

Explanation:
1. Take the motorcycle direction as positive, so \(v_M=+80\) and \(v_C=-65\) km/h
2. \(v_{MC}=v_M-v_C=80-(-65)=145\) km/h

Teacher's Note:
1. For opposite directions the relative speed is the sum of the speeds.
2. Always subtract the velocities as signed quantities.

Answer: (a) 23 m/s towards east

Explanation:
1. Both velocities are towards east, so they add
2. \(v=20+3=23\) m/s towards east

Teacher's Note:
1. Velocity relative to ground = velocity relative to train + velocity of the train.
2. Same direction gives addition.

Answer: (c) 1100 km/h

Explanation:
1. The gases are ejected backwards, opposite to the plane's motion
2. Taking the plane's direction as positive: \(v=500-1600=-1100\) km/h
3. The speed relative to the ground is 1100 km/h

Teacher's Note:
1. Exhaust moves opposite to the aircraft, so the velocities subtract.
2. Speed is the magnitude of the resulting velocity.

Answer: (a) 5 m/s due north

Explanation:
1. Train: \(54\) km/h \(=15\) m/s north; monkey relative to the train: \(36\) km/h \(=10\) m/s south
2. \(v=15-10=5\) m/s towards north

Teacher's Note:
1. Convert km/h to m/s by multiplying by 5/18.
2. Running against the train's motion subtracts the speeds.

Answer: (c) g + a, g

Explanation:
1. Relative to the ground the ball is in free fall, so its acceleration is g downwards
2. The lift accelerates upwards with a, so relative to the lift the ball has \(g+a\) downwards

Teacher's Note:
1. An observer in an accelerating frame sees the extra acceleration of the frame.
2. The ground observer sees only gravity.

Answer: (b) 55 km/h

Explanation:
1. Relative speed \(=\dfrac{100}{6}\) m/s \(=\dfrac{100}{6}\times\dfrac{18}{5}=60\) km/h
2. Opposite directions: \(v+5=60\)
3. \(v=55\) km/h

Teacher's Note:
1. Distance covered while crossing equals the length of the train.
2. Opposite directions add the speeds.

Answer: (d) 12 s

Explanation:
1. Relative speed \(=20+5=25\) m/s
2. \(t=\dfrac{300}{25}=12\) s

Teacher's Note:
1. The bird covers the length of the train relative to the train.
2. Opposite motion means the speeds add.

Answer: (b) 4 s

Explanation:
1. Relative speed \(=35+15=50\) m/s
2. Total distance to be covered \(=100+100=200\) m
3. \(t=\dfrac{200}{50}=4\) s

Teacher's Note:
1. When two trains cross, the distance covered is the sum of their lengths.
2. Use the relative speed for opposite directions.

Answer: (c) 12.5 s

Explanation:
1. The train enters when its front reaches the bridge and leaves when its rear leaves it, so the distance covered is \(200+300=500\) m
2. Average speed \(=\dfrac{30+50}{2}=40\) m/s for uniform acceleration
3. \(t=\dfrac{500}{40}=12.5\) s

Teacher's Note:
1. Crossing a bridge means covering bridge length plus train length.
2. Uniform acceleration allows using the mean of the initial and final speeds.

Answer: (b) 4.4 m/s and 3.6 m/s

Explanation:
1. Let the speeds be \(u\) and \(v\) with \(u>v\)
2. Opposite directions: \(u+v=8\); same direction: \(u-v=0.8\)
3. \(u=4.4\) m/s and \(v=3.6\) m/s

Teacher's Note:
1. Relative speed is the sum for opposite directions and the difference for the same direction.
2. Solve the two simple equations together.

Answer: (d) 2 m/s

Explanation:
1. Speed relative to the bank \(=\dfrac{30}{10}=3\) m/s
2. Upstream: \(5-v_r=3\)
3. \(v_r=2\) m/s

Teacher's Note:
1. Upstream swimming speed equals swimmer speed minus river speed.
2. Find the speed relative to the ground first.

Answer: (d) 25.71 s

Explanation:
1. Let the length be L. Walking speed \(=\dfrac L{45}\), escalator speed \(=\dfrac L{60}\)
2. Combined speed \(=\dfrac L{45}+\dfrac L{60}=\dfrac{7L}{180}\)
3. \(t=\dfrac{180}{7}\approx25.71\) s

Teacher's Note:
1. When both motions help, their speeds add; the reciprocals of times add.
2. A quick formula is \(\dfrac1t=\dfrac1{t_1}+\dfrac1{t_2}\).

Answer: (a) 8 s

Explanation:
1. The cyclist covers \(20t\) and the bus covers \(\tfrac12(2)t^2=t^2\), with a 96 m head start for the bus
2. \(20t=96+t^2\Rightarrow t^2-20t+96=0\)
3. \(t=8\) s or 12 s; the first meeting, at 8 s, is the overtaking (the bus catches up again at 12 s)

Teacher's Note:
1. Write both position equations from the same origin and equate them.
2. Take the smaller root as the first overtaking.

Answer: (c) 160 min

Explanation:
1. Still water: \(\dfrac{20}{2}\) gives boat speed 10 km/h
2. Upstream speed 5 km/h takes 2 h; downstream speed 15 km/h takes \(\dfrac{10}{15}=\dfrac23\) h
3. Total \(=\dfrac83\) h \(=160\) min

Teacher's Note:
1. Upstream speed is boat speed minus river speed and downstream is the sum.
2. The trip takes longer in a flowing river than in still water.

Answer: (b) 45 min

Explanation:
1. Upstream speed \(=6-2=4\) km/h, time \(=\dfrac24=0.5\) h
2. Downstream speed \(=6+2=8\) km/h, time \(=\dfrac28=0.25\) h
3. Total \(=0.75\) h \(=45\) min (the width of the river does not matter)

Teacher's Note:
1. The distance is measured along the flow, so the river width is irrelevant.
2. Add the upstream and downstream times.