Extra 30 detailed short-answer questions answers from the Motion chapter:8 of Class 9 CBSE Science:-
Differentiate between distance and displacement with suitable examples.
Answer:
Distance: It is the total length of the path covered by an object, irrespective of direction. It is a scalar quantity and always positive.
Displacement: It is the shortest straight-line distance between the initial and final position of the object. It is a vector quantity and can be positive, negative, or zero.
Example:
A person walks 4 m east and then 3 m west.
Distance = 4 + 3 = 7 m.
Displacement = 1 m east (shortest distance).
What is uniform motion? How is it different from non-uniform motion? Give examples.
Answer:
Uniform Motion: When an object covers equal distances in equal intervals of time, it is in uniform motion.
Example: A car moving at 60 km/h on a straight road.
Non-Uniform Motion: When an object covers unequal distances in equal intervals of time, it is in non-uniform motion.
Example: A bus moving in city traffic, where speed keeps changing.
Define speed and velocity. Differentiate between them.
Answer:
Speed: Distance travelled per unit time. It is a scalar quantity. Speed=Distance Time\text{Speed} = \frac{\text{Distance}}{\text{Time}}Speed=Time Distance
Velocity: Displacement per unit time. It is a vector quantity and has direction. Velocity=Displacement Time\text{Velocity} = \frac{\text{Displacement}}{\text{Time}}Velocity=TimeDisplacement
Differences:
| Speed | Velocity |
|--------|-----------|
| Scalar quantity | Vector quantity |
| Always positive | Can be positive, negative, or zero |
| Does not depend on direction | Depends on direction |
| Example: 40 km/h | Example: 40 km/h east |
Explain acceleration with its SI unit and formula. Give an example.
Answer:
Acceleration: It is the rate of change of velocity.
Formula: a=v−uta = \frac{v - u}{t}a=tv−u, where v is final velocity, u is initial velocity, and t is time.
SI Unit: m/s² (meter per second squared).
Example: A car increases its speed from 10 m/s to 30 m/s in 5 seconds.
a=30−105=4a = \frac{30 - 10}{5} = 4a=530−10=4 m/s²
Define and explain retardation with an example.
Answer:
Retardation (Negative Acceleration): It is the decrease in velocity per unit time.
Formula: a=v−uta = \frac{v - u}{t}a=tv−u (where v<uv < uv<u).
Example: A car moving at 30 m/s slows down to 10 m/s in 5 s.
a=10−305=−4a = \frac{10 - 30}{5} = -4a=510−30=−4 m/s² (negative sign indicates retardation).
What is meant by uniform acceleration? Give an example and its velocity-time graph.
Answer:
Uniform Acceleration: When an object's velocity changes by equal amounts in equal intervals of time.
Example: A free-falling object under gravity (g=9.8g = 9.8g=9.8 m/s²).
The velocity-time graph for uniform acceleration is a straight line sloping upwards.
A car starts from rest and attains a velocity of 20 m/s in 4 seconds. Find its acceleration.
Answer:
Given: u = 0, v = 20 m/s, t = 4 s
Using formula a=v−uta = \frac{v - u}{t}a=tv−u,
a=20−04=5a = \frac{20 - 0}{4} = 5a=420−0=5 m/s²
State the three equations of motion and explain their significance.
Answer:
First Equation: v=u+atv = u + atv=u+at → Relates velocity, time, and acceleration.
Second Equation: s=ut+12at2s = ut + \frac{1}{2} at^2s=ut+21at2 → Gives displacement when acceleration is uniform.
Third Equation: v2=u2+2asv^2 = u^2 + 2asv2=u2+2as → Relates velocity and displacement.
A ball is dropped from a height. How long will it take to reach the ground if the height is 78.4 m? (g = 9.8 m/s²)
Answer:
Using s=12gt2s = \frac{1}{2} g t^2s=21gt2,
78.4=12(9.8)t278.4 = \frac{1}{2} (9.8) t^278.4=21(9.8)t2,
t2=16t^2 = 16t2=16, so t=4t = 4t=4 s.
Explain free fall and its characteristics.
Answer:
A body falls under the influence of gravity alone, without resistance.
It has uniform acceleration g=9.8g = 9.8g=9.8 m/s².
Velocity keeps increasing, but mass does not affect the motion.
11. Explain relative motion with an example.
Answer:
Relative motion refers to the motion of an object with respect to another moving object.
Example:
If a person is sitting in a moving train, they appear at rest to another passenger but in motion to someone outside the train.
If Train A moves at 60 km/h and Train B at 40 km/h in the same direction, the relative speed of A with respect to B is: 60−40=20 km/h60 - 40 = 20 \text{ km/h}60−40=20 km/h
12. What is the significance of displacement being zero but distance being nonzero?
Answer:
Displacement can be zero even if distance is not.
Example: A person walks 10 meters east and then 10 meters west.
Distance travelled = 10+10=2010 + 10 = 2010+10=20 meters.
Displacement = 0 meters (since final and initial positions are the same).
This concept is important in circular motion where an object completes a full circle and returns to the starting point.
13. Explain the difference between instantaneous speed and average speed.
Answer:
Instantaneous Speed:
Speed of an object at a specific instant of time.
Example: The reading on a car’s speedometer at a given moment.
Average Speed:
Total distance traveled divided by total time taken.
Example: If a car travels 100 km in 2 hours, Average Speed=1002=50 km/h\text{Average Speed} = \frac{100}{2} = 50 \text{ km/h}Average Speed=2100=50 km/h
15. A body moves with constant velocity. What is its acceleration? Explain why.
Answer:
Acceleration is the rate of change of velocity.
If velocity is constant, there is no change in velocity.
Hence, acceleration = 0 m/s².
Example: A car moving at 60 km/h without changing speed has zero acceleration.
16. How can a body have zero displacement but nonzero distance?
Answer:
If an object returns to its starting position, displacement is zero, but distance is not.
Example: A person jogging around a circular track and returning to the starting point.
Distance = Length of path covered.
Displacement = 0, since the start and end points are the same.
17. Explain why velocity-time graphs are important in motion analysis.
Answer:
Velocity-time graphs provide useful information about an object’s motion.
Key Features:
Slope of the graph = Acceleration.
Area under the graph = Displacement.
Example:
A horizontal line represents constant velocity.
An upward sloping line shows increasing velocity (acceleration).
A downward sloping line represents retardation (negative acceleration).
18. How does acceleration change in free fall?
Answer:
In free fall, an object falls under the influence of gravity alone.
The acceleration due to gravity (ggg) is constant at 9.8 m/s² downward.
Velocity increases by 9.8 m/s every second.
Example:
If an object falls for 3 seconds, v=u+gt=0+(9.8×3)=29.4 m/sv = u + gt = 0 + (9.8 \times 3) = 29.4 \text{ m/s}v=u+gt=0+(9.8×3)=29.4 m/s
19. Why is motion in a straight line simpler than two-dimensional motion?
Answer:
One-dimensional motion (straight line):
Has only one direction (x-axis or y-axis).
Example: A car moving on a straight road.
Two-dimensional motion (plane motion):
Requires both x and y coordinates.
Example: A projectile (throwing a ball).
Straight-line motion is simpler because only one equation is needed instead of vector equations.
20. What is non-uniform acceleration? Give an example.
Answer:
Non-uniform acceleration occurs when velocity changes unequally over time.
Example:
A car moving in traffic where speed increases and decreases randomly.
A falling parachute that decelerates due to air resistance.
In such cases, acceleration varies at different moments.
