Class 11 Physics · Original EZT practice
Motion in a Plane (Class 11 Chapter 3): summary and practice
One-dimensional motion grows up here: velocity and acceleration become vectors, and two short formulas for a projectile carry most of the chapter's numericals. All questions are original EZT tutor practice following the chapter's own section order.
8 original worked examples, 8 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 20 October 2026:
1. Scalars and vectors
Scalars carry only magnitude; vectors carry magnitude and direction and add by the triangle or parallelogram law. Position, displacement, velocity and acceleration are all vectors in plane motion.
2. Vector arithmetic
Resolve a vector into components along the axes: vx = v cos theta, vy = v sin theta. Add component by component, then rebuild the magnitude with Pythagoras and the direction with tan.
3. Projectile motion
With launch speed v0 at angle theta: time of flight is 2 v0 sin theta / g, maximum height is v0^2 sin^2 theta / 2g, and range is v0^2 sin 2theta / g. Complementary launch angles give the same range.
4. Uniform circular motion
Constant speed around a circle still means acceleration: a = v^2/R, directed toward the centre. The velocity vector keeps turning even though its length does not change.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Tell scalars from vectors
Choose the method: Ask whether direction changes the answer.
EZT practice question
Sort these into scalars and vectors: distance, displacement, speed, velocity, acceleration.
Show the worked solution
- Distance and speed have no direction - scalars.
- Displacement, velocity and acceleration need direction - vectors.
Answer: Scalars: distance, speed. Vectors: displacement, velocity, acceleration.
Check: An object can have constant speed with changing velocity - only possible because velocity is a vector.
Watch for this: Do not call speed and velocity interchangeable; only velocity carries direction.
Now try it yourself
Is the magnitude of a vector a scalar?
Check your answer
Yes.
Magnitude is just a length - no direction left.
Add vectors componentwise
Choose the method: Add the i parts and the j parts separately.
EZT practice question
A = 3i + 4j and B = i - 2j. Find A + B, |A| and |A + B|.
Show the worked solution
- A + B = (3 + 1)i + (4 - 2)j = 4i + 2j.
- |A| = sqrt(9 + 16) = 5.
- |A + B| = sqrt(16 + 4) = sqrt20, about 4.47.
Answer: A + B = 4i + 2j; |A| = 5; |A + B| = about 4.47.
Check: The magnitude of a sum is not the sum of magnitudes: 4.47 is not 5 + |B|.
Watch for this: Do not add the magnitudes 5 and sqrt5; add components instead.
Now try it yourself
Add (2i + j) and (i + 3j), and find the magnitude.
Check your answer
3i + 4j, magnitude 5.
sqrt(9 + 16) = 5.
Resolve a vector into components
Choose the method: vx = v cos theta along the angle, vy = v sin theta across it.
EZT practice question
A projectile launches at 20 m/s at 30 degrees above the horizontal. Find its initial velocity components.
Show the worked solution
- vx = 20 cos 30 = 20 x 0.866, about 17.3 m/s.
- vy = 20 sin 30 = 10 m/s.
Answer: vx about 17.3 m/s and vy = 10 m/s.
Check: sqrt(17.3^2 + 10^2) is about 20 - the original speed.
Watch for this: Do not swap sin and cos; the component along the angle uses cos.
Now try it yourself
Same speed at 60 degrees - components?
Check your answer
vx = 10 m/s, vy about 17.3 m/s.
The components swap because 30 and 60 degrees are complementary.
Run a full projectile numerical
Choose the method: Range, height and time of flight from the three formulas.
EZT practice question
A projectile launches at 20 m/s at 30 degrees with g = 9.8 m/s^2. Find the range, maximum height and time of flight.
Show the worked solution
- R = v0^2 sin 2theta/g = 400 x sin60/9.8 = 400 x 0.866/9.8, about 35.35 m.
- H = v0^2 sin^2 theta/2g = 400 x 0.25/19.6, about 5.10 m.
- T = 2 v0 sin theta/g = 20/9.8, about 2.04 s.
Answer: Range about 35.35 m, maximum height about 5.10 m, time of flight about 2.04 s.
Check: Height over range is 5.10/35.35, close to (1/4) tan30 = 0.144 - consistent.
Watch for this: Use sin 2theta for range, not sin theta; the two formulas differ.
Now try it yourself
What is the time of flight for the same projectile launched at 60 degrees?
Check your answer
About 3.54 s.
T = 2 x 20 x sin60/9.8 = 34.64/9.8.
Use the complementary-angle range rule
Choose the method: sin 2theta is equal for theta and 90 - theta.
EZT practice question
The 20 m/s projectile at 30 degrees had a range of about 35.35 m. What is the range at 60 degrees?
Show the worked solution
- sin(2 x 60) = sin 120 = sin 60.
- Same v0 and same sin 2theta, so the same range.
Answer: About 35.35 m - complementary angles give equal ranges.
Check: The 60 degree shot goes higher and stays up longer, but lands just as far.
Watch for this: Do not assume the steeper shot goes farther; equal range, different arc.
Now try it yourself
Which launch angle gives the maximum range?
Check your answer
45 degrees.
sin 2theta peaks at 2theta = 90 degrees, so theta = 45 degrees.
Compute centripetal acceleration
Choose the method: a = v^2/R toward the centre.
EZT practice question
A particle moves at 10 m/s around a circle of radius 5 m. Find its acceleration.
Show the worked solution
- a = v^2/R.
- a = 100/5 = 20 m/s^2.
Answer: 20 m/s^2, directed toward the centre.
Check: The speed is constant but the direction changes - so an acceleration must exist.
Watch for this: Do not say zero acceleration because speed is constant; velocity keeps turning.
Now try it yourself
Same circle at 5 m/s - acceleration?
Check your answer
5 m/s^2.
a = 25/5 = 5 m/s^2.
Separate position vector from displacement
Choose the method: Position is measured from the origin; displacement is final minus initial.
EZT practice question
A particle moves from r1 = 2i + 3j to r2 = 5i + 7j. Find the displacement.
Show the worked solution
- Displacement = r2 - r1.
- = (5 - 2)i + (7 - 3)j.
Answer: 3i + 4j, with magnitude 5.
Check: Displacement depends only on the endpoints, not the path taken.
Watch for this: Do not add the two position vectors; displacement is their difference.
Now try it yourself
What is the magnitude of this displacement?
Check your answer
5.
sqrt(9 + 16) = 5.
Handle constant acceleration in a plane
Choose the method: The one-dimensional equations apply component by component.
EZT practice question
A body starts with velocity 3i m/s and has constant acceleration 2j m/s^2. What is its velocity after 4 s?
Show the worked solution
- v = u + at, componentwise.
- x-component stays 3; y-component grows to 0 + 2 x 4 = 8.
Answer: v = 3i + 8j m/s.
Check: The x-component is untouched because the acceleration has no x-part.
Watch for this: Do not mix components across axes; each axis follows its own one-dimensional equation.
Now try it yourself
What is the speed after 4 s?
Check your answer
sqrt(9 + 64) = sqrt73, about 8.54 m/s.
Speed is the magnitude of the velocity vector.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Range of a projectile?
Show answer and reason
R = v0^2 sin 2theta / g.
Complementary angles share a range because sin 2theta matches.
2. Time of flight?
Show answer and reason
T = 2 v0 sin theta / g.
Only the vertical component fights gravity.
3. Acceleration in uniform circular motion?
Show answer and reason
a = v^2/R, toward the centre.
Constant speed is not zero acceleration - direction keeps changing.
4. How do you add vectors analytically?
Show answer and reason
Add the components axis by axis, then rebuild magnitude and direction.
Magnitudes alone do not add.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- A projectile launches at 20 m/s at 60 degrees with g = 9.8 m/s^2. Find its range.
- Add (2i + j) and (i + 3j) and find the magnitude of the sum.
- A particle moves at 5 m/s in a circle of radius 5 m. Find its acceleration.
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Questions about this practice page
- What is the difference between a scalar and a vector?
- A scalar has only magnitude (distance, speed); a vector has magnitude and direction (displacement, velocity, acceleration) and adds by component.
- Why do complementary launch angles give the same range?
- Range depends on sin 2theta, and sin of 2theta equals sin of 2(90 - theta) - so 30 and 60 degrees land equally far.
- What is the direction of acceleration in uniform circular motion?
- Toward the centre of the circle: a = v^2/R inward, even though the speed stays constant.
- How do I resolve a vector?
- Multiply the magnitude by cos theta for the component along the angle and by sin theta for the component across it.
- Are these the NCERT exercise questions for Motion in a Plane (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.