JEE Physics Motion In A Straight Line MCQs Set 26

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

Explanation:
1. In uniform motion velocity is constant, so average and instantaneous velocities are equal and the Assertion is true.
2. The Reason is false because velocity does not change.

Teacher's Note:
1. Uniform velocity means constant velocity.

Answer: (b) Displacement

Explanation:
1. Distance covered is always positive for a moving particle, so average speed is also positive.
2. Displacement is zero if the particle returns to its starting point.

Teacher's Note:
1. Displacement is a vector and can cancel out; distance cannot.

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

Explanation:
1. Speed is a scalar defined as distance per unit time, so it is never negative.
2. The Reason correctly explains why.

Teacher's Note:
1. Velocity can be negative; speed cannot.

Answer: (d) remains constant

Explanation:
1. Speed = distance/time. If distance ∝ time, the ratio is constant, so the speed is constant.

Teacher's Note:
1. A straight line through the origin on a distance-time graph means uniform speed.

Answer: (b) uniform acceleration

Explanation:
1. In vacuum only gravity acts, giving a constant acceleration \( g \).

Teacher's Note:
1. Velocity and momentum change continuously during free fall.

Answer: (d) must be zero

Explanation:
1. Zero distance means the particle never moved, so its final position equals its initial position.
2. Hence the displacement must be zero.

Teacher's Note:
1. Displacement magnitude is never larger than distance.

Answer: (b) free fall

Explanation:
1. Motion under gravity alone is free fall, with constant acceleration \( g \) near the Earth.

Teacher's Note:
1. Free fall is independent of the mass of the body (in vacuum).

Answer: (c) D = S

Explanation:
1. With constant velocity the direction never changes, so the path is a straight line without reversal.
2. Hence the magnitude of displacement equals the distance.

Teacher's Note:
1. Equality of distance and displacement requires motion in one direction.

Answer: (b) 4 s

Explanation:
1. Relative speed = \( 25 m/s \).
2. Total length = \( 100 m \), so \( t = 100/25 = 4 s \).

Teacher's Note:
1. Add both lengths and both speeds.

Answer: (a) t²

Explanation:
1. \( dx/dt = \sqrt{x} \) gives \( dx/\sqrt{x} = dt \), so \( 2\sqrt{x} = t \).
2. \( x = t^{2}/4 \), hence \( x ∝ t^{2} \).

Teacher's Note:
1. Separate the variables and integrate.

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

Explanation:
1. \( h \) = ut − ½gt², so ½gt² − ut \( + h = 0 \) has roots \( t_{1} \) and \( t_{2} \).
2. The product of roots \( t_{1}t_{2} = 2h/g \), so \( h = \frac{1}{2} g t_{1}t_{2} \).

Teacher's Note:
1. Use the product of the roots of the quadratic in \( t \).

Answer: (a) 11 m

Explanation:
1. Distance = \( 2 + 4 + 5 = 11 m \).
2. Distance is a scalar and is never negative.

Teacher's Note:
1. Add all the path lengths to find the distance.

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

Explanation:
1. At the turning point the velocity is zero momentarily, so the Assertion is true.
2. The body can still have acceleration, so the Reason is false.

Teacher's Note:
1. Zero velocity with non-zero acceleration is possible.

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

Explanation:
1. In free fall acceleration is \( g \), so the velocity cannot stay constant and the Assertion is false.
2. The Reason is true.

Teacher's Note:
1. Constant velocity requires zero acceleration.

Answer: (d) Both velocity and acceleration

Explanation:
1. The direction of motion is given by the velocity and the way it changes is given by the acceleration.
2. Together they decide whether the path is a straight line, a parabola, a circle and so on.

Teacher's Note:
1. Velocity and acceleration being parallel gives straight-line motion.

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

Explanation:
1. Displacement is the area under \( v-t \), velocity is the slope of the \( x-t \) graph, acceleration is the slope of the \( v-t \) graph and instantaneous velocity is the slope of the tangent to \( x-t \).

Teacher's Note:
1. Slope goes one step down, area one step up the chain \( x, v, a \).

Answer: (c) \( (2\hat i-2\hat j)\ \text{m} \)

Explanation:
1. Displacement = final − initial = \( (4 - 2) i + (1 - 3) j = 2i - 2j \).

Teacher's Note:
1. Always subtract the initial position from the final position.

Answer: (b) The speed is, in general, less than the magnitude of the velocity

Explanation:
1. Speed (average) is greater than or equal to the magnitude of average velocity, never less.
2. The other statements are correct.

Teacher's Note:
1. Distance cannot be less than the magnitude of displacement.

Answer: (d) change in velocity

Explanation:
1. \( \Deltav \) = ∫ a \( dt \), which is the area under the \( a-t \) graph.

Teacher's Note:
1. Remember: \( v-t \) area gives displacement, \( a-t \) area gives change in velocity.

Answer: (b) I and III

Explanation:
1. I is true (highest point).
2. II is false because constant velocity means constant speed.
3. III is true (uniform circular motion).
4. IV is false because a thrown ball reverses its direction under constant \( g \).

Teacher's Note:
1. Constant acceleration does not mean constant direction of velocity.

Answer: (d) 78.4 m

Explanation:
1. The relative velocity is \( 25 m/s \), and the balls approach each other with this constant relative speed, so they meet at \( t = 100/25 = 4 s \).
2. Distance covered by the first ball = \( \frac{1}{2}(9.8)(16) = 78.4 m \).

Teacher's Note:
1. The relative acceleration is zero, so the meeting time depends only on the gap and the initial speed.

Answer: (c) both A and B strike the ground with the same velocity

Explanation:
1. A returns to the top with velocity \( v \) downward, so afterwards it moves exactly like \( B \).
2. Both reach the ground with the same speed.

Teacher's Note:
1. Use energy conservation: the speed at the ground depends only on the initial speed and the height.

Answer: (b) II and III

Explanation:
1. Constant velocity implies constant speed, so I is false.
2. II is true for circular motion and III is true at the top of a vertical throw.

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

Answer: (d) A train running on a straight track

Explanation:
1. One-dimensional motion takes place along a straight line.
2. A train on a straight track moves along one line, while the other options involve curved or two-dimensional paths.

Teacher's Note:
1. Rectilinear motion needs only one coordinate to describe the position.

Answer: (b) a is constant and not zero

Explanation:
1. The slope of the \( v-t \) graph is the acceleration.
2. A straight inclined line has constant non-zero slope, so a is constant and not zero.

Teacher's Note:
1. A horizontal line would mean \( a = 0 \).