Class 9 Physics · Chapter 2
Kinematics Revision Guide
Corrected answers and targeted practice for questions that were incorrect or only partially correct in the original exercise.
Question 6Explain the motion of a body undergoing circular motion, with an example.
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Model answer
Circular motion is motion along a circular path around a fixed centre or axis. In uniform circular motion, the speed is constant but the velocity is not, because the direction of motion changes continuously. This change in velocity means that the body has an acceleration directed towards the centre (centripetal acceleration).
Remember
A stone tied to a string and whirled in a circle; a satellite orbiting Earth; or a point on a rotating fan blade.
What to fix
The earlier response mentioned non-uniform velocity and a merry-go-round, but did not clearly explain the circular path or the continuous change in direction.
Practice check
A car travels around a circular track at constant speed. Is its velocity constant? Explain.
Question 8Define time of ascent and time of descent.
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Model answer
Time of ascent is the time taken by a body projected upward to travel from its point of projection to its maximum height. Time of descent is the time taken to travel from the maximum height back to the ground or the original level.
Remember
For a body that returns to its launch level without air resistance: time of ascent = time of descent = u/g.
What to fix
The definition of descent incorrectly described reaching maximum height; that is ascent.
Practice check
A ball takes 3 s to reach maximum height. How long will it take to return to the same level?
Question 10A body projected at an angle other than 90 degrees to the horizontal is known as what?
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Model answer
It is called an oblique projectile. Its initial velocity has both horizontal and vertical components, and its trajectory is parabolic when air resistance is ignored.
Remember
The word 'projectile' is too general; 'oblique projectile' is the precise term required.
What to fix
The earlier answer omitted the word 'oblique'.
Practice check
Classify a ball projected at 35 degrees above the horizontal.
Question 11What are scalar and vector quantities? Give examples.
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Model answer
A scalar quantity has magnitude only. A vector quantity has both magnitude and direction.
Remember
Scalars: distance, speed, mass, time. Vectors: displacement, velocity, acceleration, force.
What to fix
Saying only that scalars do not require direction is incomplete unless magnitude is also stated.
Practice check
Classify each as scalar or vector: 12 m, 12 m east, 5 m/s, 5 m/s north.
Question 13Give the CGS and MKS units of distance and displacement.
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Model answer
Both distance and displacement use centimetre (cm) in the CGS system and metre (m) in the MKS system.
Remember
CGS = centimetre-gram-second; MKS = metre-kilogram-second.
What to fix
The MKS unit was left uncertain. It is metre (m).
Practice check
Convert a displacement of 2.4 m into centimetres.
Question 16Why is distance scalar while displacement is vector?
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Model answer
Distance is the total length of the actual path travelled. It has magnitude but no direction, so it is scalar. Displacement is the straight-line change from the initial position to the final position and includes direction, so it is vector.
Remember
After one complete lap of a 400 m track: distance = 400 m, displacement = 0 m.
What to fix
The earlier response reversed the ideas of actual path and shortest straight-line separation.
Practice check
A student walks 3 m east and then 3 m west. Find distance and displacement.
Question 22What are acceleration and retardation?
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Model answer
Acceleration is the rate of change of velocity with time: a = (v - u)/t. Retardation (deceleration) is acceleration that acts opposite to the velocity and reduces speed.
Remember
The sign of acceleration depends on the selected positive direction. Acceleration is not always positive, and retardation is not defined merely as negative acceleration.
What to fix
The earlier response treated positive and negative signs as absolute definitions.
Practice check
A car slows from 20 m/s to 8 m/s in 4 s. Calculate its acceleration.
Question 26A particle moves P to Q at velocity v1 and Q to P at velocity v2. What is its average velocity for the complete trip?
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Model answer
Average velocity = total displacement / total time. The particle returns to P, so its total displacement is zero. Therefore its average velocity is 0, regardless of v1 and v2.
Remember
Do not confuse average velocity with average speed. The average speed would generally be non-zero.
What to fix
This question was left blank.
Practice check
A runner completes one full lap and returns to the starting point. What is the average velocity?
Question 28A body travels at v1 for the first half of the total time and at v2 for the remaining half. Find its average velocity.
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Model answer
For equal time intervals, average velocity is the arithmetic mean: average velocity = (v1 + v2)/2, provided both velocities are along the stated straight-line direction.
Remember
If the question instead specified equal distances, the result would be 2v1v2/(v1 + v2).
What to fix
This question was left blank.
Practice check
A body moves at 10 m/s for 4 s and 18 m/s for the next 4 s. Find its average velocity.
Question 30What is a trajectory?
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Model answer
A trajectory is the path followed by a moving body or particle through space.
Remember
For an oblique projectile under gravity, ignoring air resistance, the trajectory is a parabola. However, not every trajectory is parabolic.
What to fix
The earlier response described a condition for a parabolic projectile path instead of defining trajectory generally.
Practice check
State the trajectory of (a) an oblique projectile and (b) a satellite in a circular orbit.
Question 31Derive the maximum height reached by a body projected vertically upward.
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Model answer
Take upward as positive. For upward motion, a = -g. Use v^2 = u^2 + 2as. At maximum height, v = 0 and s = H. Therefore 0 = u^2 - 2gH, so H = u^2/(2g).
Remember
The result must be accompanied by the equation used, the sign of acceleration, and the condition v = 0 at maximum height.
What to fix
Only the final formula was supplied; the requested derivation was missing.
Practice check
Find the maximum height for u = 14 m/s and g = 9.8 m/s^2.
Question 33What are the different types of motion? Give an example of each.
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Model answer
Translational motion: the body moves from one position to another. Rectilinear example: a train on a straight track. Curvilinear example: a thrown ball. Rotational motion: a body turns about an axis, such as a ceiling fan. Oscillatory motion: repeated to-and-fro motion about a mean position, such as a pendulum.
Remember
Uniform and non-uniform describe how motion changes; they are not substitutes for the main path/behaviour categories requested here.
What to fix
The earlier list contained some valid terms but no examples and mixed categories.
Practice check
Name the type of motion: a lift moving vertically, a spinning wheel, and a playground swing.
Question 34Derive the time taken by a vertically projected body to reach maximum height.
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Model answer
For upward motion, v = u - gt. At maximum height, v = 0. Hence 0 = u - gt, giving time of ascent t = u/g.
Remember
The maximum-height formula H = u^2/(2g) answers a different question.
What to fix
The earlier answer gave maximum height rather than time of ascent.
Practice check
A ball is projected upward at 19.6 m/s. Find its time of ascent.
Question 35A runner covers 200 cm in the 2nd second and 220 cm in the 4th second. Find initial velocity and uniform acceleration.
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Model answer
Use distance in the nth second: S_n = u + (a/2)(2n - 1). Convert 200 cm = 2.00 m and 220 cm = 2.20 m. For n=2: 2.00 = u + 1.5a. For n=4: 2.20 = u + 3.5a. Subtraction gives 0.20 = 2a, so a = 0.10 m/s^2. Then u = 2.00 - 1.5(0.10) = 1.85 m/s.
Remember
Final answer: initial velocity = 1.85 m/s; acceleration = 0.10 m/s^2.
What to fix
The method and initial velocity were correct, but acceleration was written as 0.01 m/s^2.
Practice check
Check the result by substituting u and a into the expression for the 4th second.
Question 42A train travels between two stations at 40 km/h and returns at 60 km/h. Find average speed and average velocity.
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Model answer
For equal one-way distances, average speed = 2v1v2/(v1 + v2) = 2(40)(60)/(40+60) = 48 km/h. For the complete return journey, total displacement is zero, so average velocity = 0 km/h.
Remember
Do not take (40+60)/2 because the train spends unequal times at the two speeds.
What to fix
Average velocity was correct, but average speed was written as 50 km/h.
Practice check
A car travels equal distances at 30 km/h and 50 km/h. Find its average speed.
Question 43Use the velocity-time graph to find (i) velocity of projection and (ii) maximum height.
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Model answer
The graph starts at +20 m/s, so the velocity of projection is 20 m/s. Maximum height equals displacement during ascent, which is the positive area under the velocity-time graph from 0 to 2 s: H = 1/2 x 2 s x 20 m/s = 20 m.
Remember
Area under a velocity-time graph gives displacement. Only the portion above the time axis represents upward displacement.
What to fix
Part (i) was correct. Part (ii) identified the method but did not finish the calculation.
Practice check
If the same graph crossed the time axis at 3 s with initial velocity 24 m/s, find the maximum height.
Question 44Determine acceleration: (a) constant speed 20 m/s for 12 s; (b) velocity changes from 0 m/h to 360 m/min in 4.2 s.
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Model answer
(a) Constant velocity gives a = 0 m/s^2. (b) Convert 360 m/min to 6 m/s. Then a = (v-u)/t = (6-0)/4.2 = 1.43 m/s^2 (approximately).
Remember
Always convert velocities to consistent units before applying a = (v-u)/t.
What to fix
Part (a) and the conversion in part (b) were correct, but the final division and acceleration unit were not completed.
Practice check
A bicycle speeds up from 2 m/s to 11 m/s in 6 s. Find its acceleration.
Practice Answer Key
- Q6. No. Its speed is constant, but its direction changes, so velocity changes.
- Q8. 3 s.
- Q10. An oblique projectile.
- Q11. Scalar, vector, scalar, vector.
- Q13. 240 cm.
- Q16. Distance = 6 m; displacement = 0 m.
- Q22. -3 m/s^2 (taking forward as positive).
- Q26. 0.
- Q28. 14 m/s.
- Q30. Parabolic; circular.
- Q31. 10 m.
- Q33. Rectilinear translational; rotational; oscillatory.
- Q34. 2 s.
- Q35. 2.20 = 1.85 + 3.5(0.10) = 2.20 m.
- Q42. 37.5 km/h.
- Q43. 36 m.
- Q44. 1.5 m/s^2.
