JEE Physics Motion In A Straight Line MCQs Set 25

Answer: (b) 735 m

Explanation:
1. After \( 5 s \): \( v = 19.6(5) = 98 m/s \) and \( h_{1} = \frac{1}{2}(19.6)(25) = 245 m \).
2. Afterwards \( h_{2} = v^{2}/(2g) = 9604/19.6 = 490 m \).
3. Total = \( 245 + 490 = 735 m \).

Teacher's Note:
1. Treat the powered and the coasting phases separately.

Answer: (a) an acceleration

Explanation:
1. The coin keeps the horizontal velocity it had at the toss.
2. If the train speeds up, the passenger moves ahead of the coin, so the coin lands behind him.

Teacher's Note:
1. With uniform speed the coin would return to his hand.

Answer: (a) \( (velocity)^{3/2} \)

Explanation:
1. \( v = dx/dt = -(t + 5)^{-2} \) and \( a = 2(t + 5)^{-3} \).
2. \( |v|^{3/2} = (t + 5)^{-3} \), so a \( ∝ v^{3/2} \).

Teacher's Note:
1. Differentiate twice and compare powers of \( (t + 5) \).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. With final velocity 0: \( s = u^{2}/(2a) \), so \( s ∝ u^{2} \).
2. The Reason gives the relation, so it explains the Assertion.

Teacher's Note:
1. Braking distance quadruples if the speed doubles.

Answer: (d) I, II and III

Explanation:
1. At the highest point velocity is zero (minimum) while acceleration is \( g \) downwards.
2. All three statements are wrong.

Teacher's Note:
1. Velocity is minimum at the top; acceleration stays at \( g \).

Answer: (c) 1 : 1

Explanation:
1. In free fall the time depends only on the height and \( g \), not on the mass.
2. So the ratio is 1 : 1.

Teacher's Note:
1. All bodies fall with the same acceleration in vacuum.

Answer: (b) \( a=\frac{dv}{dt} \)

Explanation:
1. Instantaneous acceleration is the limit of \( \Deltav/\Deltat \) as \( \Deltat \) tends to zero, which is dv/dt.

Teacher's Note:
1. Option \( (a) \) is average acceleration.

Answer: (a) 5 s

Explanation:
1. The packet starts with the balloon's velocity, \( 12 m/s \) upwards.
2. \( -65 = 12t - 5t^{2} \), so \( 5t^{2} - 12t - 65 = 0 \).
3. \( t = (12 + 38)/10 = 5 s \).

Teacher's Note:
1. Initial velocity of a released object equals the velocity of the carrier.

Answer: (a) zero

Explanation:
1. He returns to the starting point, so the displacement is zero.
2. Average velocity = 0.

Teacher's Note:
1. Average speed is 2 km/h but average velocity is zero.

Answer: (c) A is true, but R is false.

Explanation:
1. In the first \( 2 s \): \( \frac{1}{2}a(4) = 2a \). In the first \( 4 s \): \( \frac{1}{2}a(16) = 8a \), so the next \( 2 s \) give \( 6a = 3(2a) \). The Assertion is true.
2. Displacement is proportional to \( t^{2} \), not to velocity, so the Reason is false.

Teacher's Note:
1. Distances in equal successive intervals from rest are in the ratio 1 : 3 : 5.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The passenger just catches the bus if the relative velocity becomes zero as the gap closes, so for relative motion \( u^{2} \) = 2ad, giving \( u \) = √(2ad).
2. The Reason is the equation used.

Teacher's Note:
1. Work in the frame of the bus.

Answer: (c) II and III

Explanation:
1. I is correct.
2. II is wrong because average speed can exceed |average velocity|.
3. III is wrong because non-zero speed always gives a non-zero distance.
4. IV is correct (\( e.g \). a ball at its highest point).

Teacher's Note:
1. Average speed is zero only if the particle never moves.

Answer: (d) 16 m

Explanation:
1. \( v = 12 - 3t^{2} = 0 \) at \( t = 2 s \).
2. \( x(0) = 40 m \) and \( x(2) = 40 + 24 - 8 = 56 m \).
3. Distance travelled = \( 56 - 40 = 16 m \).

Teacher's Note:
1. Find the time at which velocity is zero first.

Answer: (c) zero

Explanation:
1. For uniform velocity the \( v-t \) graph is a horizontal line.
2. Its slope is the acceleration, which is zero.

Teacher's Note:
1. Slope of \( v-t \) is acceleration.

Answer: (d) 1 m/s

Explanation:
1. The difference of distances in equal successive intervals = \( a t^{2} \): \( 30 - 12 = a(9) \), so \( a = 2 m/s^{2} \).
2. \( 12 = u(3) + \frac{1}{2}(2)(9) \), so \( u = 1 m/s \).

Teacher's Note:
1. Use \( \Deltas = a t^{2} \) for successive equal intervals.

Answer: (c) A is true, but R is false.

Explanation:
1. Uniform motion in the negative direction gives a negative slope, so the Assertion is true.
2. A decreasing speed makes the graph curve towards a flatter slope, not necessarily negative, so the Reason is false.

Teacher's Note:
1. Sign of the slope gives the direction of motion.

Answer: (b) \( \frac{\text{total path length}}{\text{total time}} \)

Explanation:
1. Average speed is the total path length (distance) travelled divided by the total time taken.

Teacher's Note:
1. Total displacement over time defines average velocity, not average speed.

Answer: (b) 1 : 3 : 5

Explanation:
1. Distance in the nth interval \( ∝ (2n - 1) \) for motion from rest.
2. The ratio is 1 : 3 : 5 : 7 and so on.

Teacher's Note:
1. Galileo's odd-number rule.

Answer: (d) A-4, B-5, C-1, D-3, E-2

Explanation:
1. cm \( s^{-2} \) is the unit of acceleration, negative acceleration is retardation, straight-line motion is linear motion, area under speed-time is distance and the velocity at the top is zero.

Teacher's Note:
1. Match quantities with their meanings and units.

Answer: (c) the slope of the velocity-time graph

Explanation:
1. \( a \) = dv/dt, the slope of the velocity-time graph.

Teacher's Note:
1. The area under \( v-t \) gives displacement.

Answer: (a) \( \frac{v_1+2v_2}{3} \)

Explanation:
1. Let the total time be \( t \). Distance = \( v_{1}(t/3) + v_{2}(2t/3) \).
2. Average velocity = \( (v_{1} + 2v_{2})/3 \).

Teacher's Note:
1. When time intervals are given, the average velocity is a weighted mean by time.

Answer: (d) straight line

Explanation:
1. \( v = u \) − gt is linear in \( t \), so the \( v-t \) graph is a straight line with slope \( -g \).

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

Answer: (a) 1 m/s²

Explanation:
1. \( v^{2} = u^{2} \) − 2as gives \( 0 = 400 - 2a(200) \).
2. \( a = 1 m/s^{2} \).

Teacher's Note:
1. Use \( v^{2} = u^{2} \) + 2as with \( v = 0 \).

Answer: (d) 10 s

Explanation:
1. Relative speed = \( 10 + 5 = 15 m/s \).
2. Time = \( 150/15 = 10 s \).

Teacher's Note:
1. Opposite directions: add the speeds.

Answer: (b) less than the time of fall

Explanation:
1. While rising, gravity and air drag both act downwards, so the retardation is greater than \( g \).
2. While falling, drag acts upwards, so acceleration is less than \( g \).
3. The same distance is covered, so the rise takes less time.

Teacher's Note:
1. With air resistance, the body returns with a speed lower than the speed of projection.