Important Questions Class 9 Science Chapter 7: Chapter 7 of Class 9 Science Motion explain the fundamental concepts of how objects move and the factors influencing their motion. Important questions cover topics like distance versus displacement, speed versus velocity and the meaning of acceleration and deceleration.
Students may need to solve problems to find average speed, read distance-time graphs, or use equations of motion for objects moving at a steady rate. It is also important to understand concepts like frame of reference and different types of motion, such as uniform and non-uniform motion. By practicing these key questions, students can strengthen their understanding of motion in the world around them.Important Questions Class 9 Science Chapter 7 PDF
Very Short Answer Questions (1 Mark)
Short Answer Questions (2 Marks)
Q.1. Distinguish between speed and velocity.
Ans: Speed is the distance traveled by a body per unit time, while velocity is the rate and direction of an object’s movement.
Q.2. Under what condition(s) is the magnitude of the average velocity of an object equal to its average speed?
Ans: The magnitude of the average velocity is equal to the average speed when the distance traveled by the body is equal to its displacement.
Q.3. What does the odometer of an automobile measure?
Ans: The odometer measures the distance covered by an automobile.
Q.4. What does the path of an object look like when it is in uniform motion?
Ans: The path of an object in uniform motion is linear, appearing as a straight line on a graph.
Q.5. Which of the following is true for displacement?
a. It cannot be zero.Ans: False; displacement can be zero if the starting and ending points are the same.
b. Its magnitude is greater than the distance traveled by the object.Ans: False; the magnitude of displacement is less than or equal to the distance traveled.
Q.6. When will you say a body is in:
i. Uniform acceleration?Ans: A body is in uniform acceleration when it travels in a straight line, and its velocity changes by equal amounts in equal intervals of time.
ii. Non-uniform acceleration?Ans: Non-uniform acceleration occurs when the velocity of an object changes by unequal amounts in equal intervals of time.
Q.7. Differentiate between Distance and Displacement
Ans: The difference between distance and displacement is as follows:
Distance | Displacement |
---|---|
1. The length of the actual path traveled by the body from the initial position to the final position. | The length of the straight line joining the initial and final positions of the body. |
2. It is a scalar quantity, meaning it has only magnitude. | It is a vector quantity, meaning it has both magnitude and direction. |
3. Distance is always positive. | Displacement can be positive, negative, or zero. |
Q.8. Define Uniform Velocity and Uniform Acceleration
Ans:
Q.9. Differentiate Between Scalars and Vectors
Ans: The difference between scalars and vectors is as follows:
Vector | Scalar |
---|---|
1. Has magnitude and specific direction. | Has magnitude but no direction. |
2. Can be positive or negative. | Always positive. |
3. Examples: Displacement, velocity. | Examples: Distance, speed. |
Short Answer Questions (3 Marks)
Q.1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Ans: Yes, an object can have zero displacement even after moving through a distance. Displacement is defined as the change in position of an object. For example, if an object travels from point A to point B and then returns to point A, the total displacement is zero because the initial and final positions are the same.
Q.2. State which of the following situations are possible and give an example for each of these:
a. An object with a constant acceleration but with zero velocity.
Ans: Yes, an object can have a constant acceleration while having zero velocity.
b. An object moving in a certain direction with acceleration in the perpendicular direction.
Ans: Yes, this situation is also possible. For instance, when an athlete runs along a circular track, their speed remains constant, but their direction changes. In this case, the acceleration is directed toward the center of the circle (centripetal acceleration) and is always perpendicular to the direction of motion. Therefore, the athlete's motion on a circular path is an example of this situation.
Q.3. A cheetah is the fastest land animal and can achieve a peak velocity of 100 km/h up to distances less than 500 m. If a cheetah spots its prey at a distance of 100 m, what is the minimum time it will take to get its prey, if the average velocity attained by it is 90 km/h?
Ans:
Average velocity = 90 km/h = 90 k m 1 h = 90 × 1000 m 60 × 60 s = 25 ms -1 Also, Average velocity = Displacement Time taken ∵ Cheetah moves in a straight line displacement is equal to 100 m. Therefore, time taken = 100 25 = 4 sQ.4. The brakes applied to a car produce an acceleration of 6 ms -2 in the opposite direction to the motion. If the car takes 2s to stop after the application of brakes, calculate the distance it travels during this time.
Ans:
We have been givenQ.5. A car starts from rest and moves along the x-axis with constant acceleration 5 ms -2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?
Ans:
Initial velocity ,u = 0 Using s = ut + 1 2 at 2 The distance travelled in first 8s, s 1 = 0 + 1 2 × 5 × 8 2 = 160 m At this point the velocity, v = u + at = 0 + 5 × 8 = 40 ms -1 So, the distance covered in last four seconds s 2 = 40 × 4 = 160 m Hence, total distance, s = s 1 + s 2 = 160 m + 160 m = 320 mQ.6. A motorcyclist drives from A to B with a uniform speed of 30 kmh -1 and returns back with a speed of 20 kmh -1 . Find its average speed.
Ans:
Q.7. An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time?
Ans:
Initial difference in height = (150 – 100) m = 50 m Distance travelled by first body in 2 s = h 1 = 0 + 1 2 g(2) 2 = 2g Distance travelled by another body in 2 s = h 2 = 0 + 1 2 g(2) 2 = 2g After 2 s, height at which the first body will be = h 1 ‘ = 150 – 2g After 2 s, height at which the second body will be = h’ 2 = 100 – 2g Thus, after 2 s, difference in height = 150 – 2g – (100 – 2g) = 50 m = initial difference in height Thus, difference in height does not vary with time.Q.13. An object starting from rest travels 20 m in first 2s and 160 m in next 4s. What will be the velocity after 7s from the start? [NCERT Exemplar]
Ans:
s 1 = ut + 1 2 at 2 or 20 = 0 + 1 2 a(2) 2 or a = 10ms -2 v = u + at = 0 + (10 × 2) = 20 ms -1 s 2 = 160 = vt’ + 1 2 a'(t’) 2 = (20 × 4) + ( 1 2 a’ × 16) ⇒ a’ = 10ms -2 Since accelerations is the same, we have v’ = 0 + (10 × 7) =70 ms -1Q.14. An electron moving with a velocity of 5 × 10 4 ms -1 enters into a uniform electric field and acquires a uniform acceleration of 104 ms -2 in the direction of its initial motion.
(i) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
(ii) How much distance the electron would cover in this time? [NCERT Exemplar]
Ans:
Given initial velocity, u = 5 × 10 4 ms -1 and acceleration, a = 10 4 ms -2 (i) Final velocity = v = 2u = 2 × 5 × 10 4 ms -1 = 10 × 10 4 ms -1 To find t, use v = u + atCBSE Class 9 Maths Syllabus | CBSE Class 9 Science Syllabus |
CBSE Class 9 Computer Application Syllabus | CBSE Class 9 Hindi Syllabus A |
CBSE Class 9 English Syllabus | CBSE Class 9 Social Science Syllabus |