CBSE Class 9 Physics Motion MCQs Set 12

Answer: (b) Displacement

Explanation:
1. Displacement depends only on the initial and final positions; they coincide, so the displacement is zero.
2. Distance, average speed and time are positive for any motion that has taken place.

Teacher's Note:
1. Displacement is a vector from start to finish.
2. Distance is the total path length.

Answer: (a) Speed

Explanation:
1. With retardation the velocity changes by -a every second, so the speed falls steadily.
2. The acceleration is constant and the displacement and distance keep increasing.

Teacher's Note:
1. Retardation means that speed falls uniformly.
2. The velocity-time graph is a falling straight line.

Answer: (c) zero

Explanation:
1. For uniform velocity the graph is a horizontal line.
2. The slope of the velocity-time graph is the acceleration, which is zero.

Teacher's Note:
1. Slope of a v-t graph = acceleration.
2. A horizontal line has zero slope.

Answer: (d) It must be zero

Explanation:
1. Zero distance means the particle has not moved at all, so its final position is its initial position.
2. Therefore the displacement must be zero.

Teacher's Note:
1. The magnitude of displacement never exceeds the distance.
2. The reverse is not true: zero displacement does not imply zero distance.

Answer: (a) 5 s

Explanation:
1. The packet leaves with u = +12 m/s (upward) and its displacement is -65 m: \( -65 = 12t - 5t^2 \).
2. \( 5t^2 - 12t - 65 = 0 \Rightarrow t = \dfrac{12 + \sqrt{144 + 1300}}{10} = \dfrac{12 + 38}{10} = 5 \) s.

Teacher's Note:
1. The packet initially moves up with the balloon's velocity.
2. Take the positive root only.

Answer: (c) 15 m/s²

Explanation:
1. Change in velocity = area under the a-t graph = area of triangle (0 to 10 s) + area of rectangle (10 to 20 s) = \( \tfrac12 \times 10 \times 20 + 10 \times 20 = 300 \) m/s.
2. Average acceleration = 300 / 20 = 15 m/s².

Teacher's Note:
1. The area under an a-t graph is the change of velocity.
2. Average acceleration = change in velocity / time.

Answer: (a) \( u + \tfrac12 at \)

Explanation:
1. Displacement \( s = ut + \tfrac12 at^2 \).
2. Average velocity = s / t = \( u + \tfrac12 at \), which equals \( \dfrac{u + v}{2} \).

Teacher's Note:
1. Average velocity = displacement / time.
2. For uniform acceleration it is the mean of u and v.

Answer: (d) 58.8 m

Explanation:
1. Taking upward as positive, the displacement is \( s = ut - \tfrac12 gt^2 = 19.6 \times 6 - 4.9 \times 36 = 117.6 - 176.4 = -58.8 \) m.
2. The ball ends 58.8 m below the starting point, so the tower is 58.8 m high.

Teacher's Note:
1. A negative displacement means the final point is below the start.
2. The ball returns to the tower's level after 4 s and then falls to the ground.

Answer: (c) 19.6 m/s

Explanation:
1. Speed after 19.6 m of free fall: \( v^2 = 2 \times 9.8 \times 19.6 = 384.16 \Rightarrow v = 19.6 \) m/s.
2. With no acceleration afterwards, the speed stays constant at 19.6 m/s till it reaches the ground.

Teacher's Note:
1. After g stops acting the body moves with uniform velocity.
2. v = 19.6 m/s is also the speed after 2 s of free fall.

Answer: (c) straight line

Explanation:
1. Velocity changes linearly with time, \( v = u + at \), which is the equation of a straight line.

Teacher's Note:
1. The slope of the line is the acceleration.
2. The displacement-time graph of uniform acceleration is a parabola.

Answer: (a) \( 2\sqrt{3} \) m towards north

Explanation:
1. East components: 5 + 4 sin 30° - 7 = 5 + 2 - 7 = 0.
2. North component: 4 cos 30° = \( 2\sqrt{3} \) m. The resultant is \( 2\sqrt{3} \) m due north.

Teacher's Note:
1. Resolve each displacement into east and north components.
2. Displacements add as vectors, not as numbers.