Class 11 Physics · Original EZT practice
Motion in a Straight Line (Class 11 Physics Chapter 2): three equations that describe every straight journey
A bus braking, a ball dropped from a roof, a train pulling away - if the path is a straight line, three small equations describe it completely. This chapter builds them from the ideas of displacement, velocity and acceleration. Below: the ideas in plain words, then original EZT practice with worked answers. (Note: this is original practice, not a textbook exercise key.)
10 original worked examples, 10 fresh self-check 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 17 October 2026:
1. Distance vs displacement, speed vs velocity
Distance is the full path travelled; displacement is the straight change in position, with direction. Speed is distance over time; velocity is displacement over time - and it can be negative or zero even when the journey was long.
2. Acceleration: change in velocity
Acceleration is the rate of change of velocity: a = (v - u)/t, in m/s2. Speeding up, slowing down (negative acceleration) and turning all count - here we stay on a straight line.
3. The three equations of motion
For uniform acceleration: v = u + at; s = ut + (1/2)at2; v2 = u2 + 2as. Pick the equation whose unknown you want from the three quantities you know.
4. Graphs that tell the story
On a position-time graph, the slope is velocity. On a velocity-time graph, the slope is acceleration and the area under the line is displacement.
5. Free fall: gravity as constant acceleration
Dropped objects fall with acceleration g = 9.8 m/s2 downward. The same three equations apply - with u = 0 for a drop, and a = -g for a toss upward.
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.
A runner jogs 400 m around a full circle back to the start. Find the dis
Choose the method: Distance is the full path travelled; displacement is the straight change in position, with direction.
EZT practice question
A runner jogs 400 m around a full circle back to the start. Find the distance and displacement.
Show the worked solution
- List the known quantities and choose the unknown you need.
- Use displacement, velocity, acceleration or a graph relation that connects those quantities.
- Keep the sign and SI unit with the result.
Answer: distance 400 m; displacement 0 - she ends where she began.
Check: The substitution and final condition match the worked statement.
Watch for this: Do not mix distance with displacement, or speed with velocity; direction and sign decide the answer.
Now try it yourself
Create one new example with different numbers and solve it by the same method.
Check your answer
Keep the final result in your notebook; the page-level Try It prompts are intentionally answer-free.
The worked answer above shows the method; independent practice should not become another answer key.
A car goes 60 km north, then 60 km south, in 2 hours. Find average speed
Choose the method: Distance is the full path travelled; displacement is the straight change in position, with direction.
EZT practice question
A car goes 60 km north, then 60 km south, in 2 hours. Find average speed and average velocity.
Show the worked solution
- List the known quantities and choose the unknown you need.
- Use displacement, velocity, acceleration or a graph relation that connects those quantities.
- Keep the sign and SI unit with the result.
Answer: average speed = 120/2 = 60 km/h; average velocity = 0/2 = 0 - net displacement is zero.
Check: The substitution and final condition match the worked statement.
Watch for this: Do not mix distance with displacement, or speed with velocity; direction and sign decide the answer.
Now try it yourself
Create one new example with different numbers and solve it by the same method.
Check your answer
Keep the final result in your notebook; the page-level Try It prompts are intentionally answer-free.
The worked answer above shows the method; independent practice should not become another answer key.
A bike goes from 10 m/s to 20 m/s in 5 s. Find its acceleration
Choose the method: Acceleration is the rate of change of velocity: a = (v - u)/t, in m/s2.
EZT practice question
A bike goes from 10 m/s to 20 m/s in 5 s. Find its acceleration.
Show the worked solution
- List the known quantities and choose the unknown you need.
- Use displacement, velocity, acceleration or a graph relation that connects those quantities.
- Keep the sign and SI unit with the result.
Answer: a = (20 - 10)/5 = 2 m/s2.
Check: The substitution and final condition match the worked statement.
Watch for this: Do not mix distance with displacement, or speed with velocity; direction and sign decide the answer.
Now try it yourself
Create one new example with different numbers and solve it by the same method.
Check your answer
Keep the final result in your notebook; the page-level Try It prompts are intentionally answer-free.
The worked answer above shows the method; independent practice should not become another answer key.
A car starts at 10 m/s and accelerates at 2 m/s2 for 5 s. Find its final
Choose the method: For uniform acceleration: v = u + at; s = ut + (1/2)at2; v2 = u2 + 2as.
EZT practice question
A car starts at 10 m/s and accelerates at 2 m/s2 for 5 s. Find its final velocity.
Show the worked solution
- List the known quantities and choose the unknown you need.
- Use displacement, velocity, acceleration or a graph relation that connects those quantities.
- Keep the sign and SI unit with the result.
Answer: v = 10 + 2 x 5 = 20 m/s.
Check: The substitution and final condition match the worked statement.
Watch for this: Do not mix distance with displacement, or speed with velocity; direction and sign decide the answer.
Now try it yourself
Create one new example with different numbers and solve it by the same method.
Check your answer
Keep the final result in your notebook; the page-level Try It prompts are intentionally answer-free.
The worked answer above shows the method; independent practice should not become another answer key.
How far does it travel in those 5 s?
Choose the method: For uniform acceleration: v = u + at; s = ut + (1/2)at2; v2 = u2 + 2as.
EZT practice question
How far does it travel in those 5 s?
Show the worked solution
- List the known quantities and choose the unknown you need.
- Use displacement, velocity, acceleration or a graph relation that connects those quantities.
- Keep the sign and SI unit with the result.
Answer: s = 10 x 5 + (1/2) x 2 x 25 = 50 + 25 = 75 m.
Check: The substitution and final condition match the worked statement.
Watch for this: Do not mix distance with displacement, or speed with velocity; direction and sign decide the answer.
Now try it yourself
Create one new example with different numbers and solve it by the same method.
Check your answer
Keep the final result in your notebook; the page-level Try It prompts are intentionally answer-free.
The worked answer above shows the method; independent practice should not become another answer key.
Check the same journey with the third equation
Choose the method: For uniform acceleration: v = u + at; s = ut + (1/2)at2; v2 = u2 + 2as.
EZT practice question
Check the same journey with the third equation.
Show the worked solution
- List the known quantities and choose the unknown you need.
- Use displacement, velocity, acceleration or a graph relation that connects those quantities.
- Keep the sign and SI unit with the result.
Answer: v2 = 100 + 2 x 2 x 75 = 400, so v = 20 m/s - matches.
Check: The substitution and final condition match the worked statement.
Watch for this: Do not mix distance with displacement, or speed with velocity; direction and sign decide the answer.
Now try it yourself
Create one new example with different numbers and solve it by the same method.
Check your answer
Keep the final result in your notebook; the page-level Try It prompts are intentionally answer-free.
The worked answer above shows the method; independent practice should not become another answer key.
A car moving at 20 m/s brakes with acceleration -5 m/s2. Find the stoppi
Choose the method: For uniform acceleration: v = u + at; s = ut + (1/2)at2; v2 = u2 + 2as.
EZT practice question
A car moving at 20 m/s brakes with acceleration -5 m/s2. Find the stopping distance.
Show the worked solution
- List the known quantities and choose the unknown you need.
- Use displacement, velocity, acceleration or a graph relation that connects those quantities.
- Keep the sign and SI unit with the result.
Answer: 0 = 400 - 10s gives s = 40 m.
Check: The substitution and final condition match the worked statement.
Watch for this: Do not mix distance with displacement, or speed with velocity; direction and sign decide the answer.
Now try it yourself
Create one new example with different numbers and solve it by the same method.
Check your answer
Keep the final result in your notebook; the page-level Try It prompts are intentionally answer-free.
The worked answer above shows the method; independent practice should not become another answer key.
A velocity-time graph is a flat line at 15 m/s from t = 0 to t = 4 s. Fi
Choose the method: On a position-time graph, the slope is velocity.
EZT practice question
A velocity-time graph is a flat line at 15 m/s from t = 0 to t = 4 s. Find the displacement.
Show the worked solution
- List the known quantities and choose the unknown you need.
- Use displacement, velocity, acceleration or a graph relation that connects those quantities.
- Keep the sign and SI unit with the result.
Answer: area = 15 x 4 = 60 m.
Check: The substitution and final condition match the worked statement.
Watch for this: Do not mix distance with displacement, or speed with velocity; direction and sign decide the answer.
Now try it yourself
Create one new example with different numbers and solve it by the same method.
Check your answer
Keep the final result in your notebook; the page-level Try It prompts are intentionally answer-free.
The worked answer above shows the method; independent practice should not become another answer key.
A stone is dropped and falls for 3 s. How far does it fall?
Choose the method: Dropped objects fall with acceleration g = 9.
EZT practice question
A stone is dropped and falls for 3 s. How far does it fall?
Show the worked solution
- List the known quantities and choose the unknown you need.
- Use displacement, velocity, acceleration or a graph relation that connects those quantities.
- Keep the sign and SI unit with the result.
Answer: s = (1/2) x 9.8 x 9 = 44.1 m.
Check: The substitution and final condition match the worked statement.
Watch for this: Do not mix distance with displacement, or speed with velocity; direction and sign decide the answer.
Now try it yourself
Create one new example with different numbers and solve it by the same method.
Check your answer
Keep the final result in your notebook; the page-level Try It prompts are intentionally answer-free.
The worked answer above shows the method; independent practice should not become another answer key.
A ball is thrown straight up at 19.6 m/s. How high does it rise?
Choose the method: Dropped objects fall with acceleration g = 9.
EZT practice question
A ball is thrown straight up at 19.6 m/s. How high does it rise?
Show the worked solution
- List the known quantities and choose the unknown you need.
- Use displacement, velocity, acceleration or a graph relation that connects those quantities.
- Keep the sign and SI unit with the result.
Answer: at the top v = 0, so 0 = (19.6)2 - 2 x 9.8 x h, giving h = 384.16/19.6 = 19.6 m. TRY IT (no answers, for practice): (a) u = 5 m/s, a = 3 m/s2, t = 4 s - find v and s. (b) A train stops from 30 m/s in 10 s - find acceleration and stopping distance. (c) What does the area under a velocity-time graph give? (d) A ball dropped from rest hits the ground at 19.6 m/s - from what height was it dropped? Edition note: the pre-rationalisation book placed this chapter third (after Physical World and Units and Measurements); the current book, with Physical World dropped, has it as Chapter 2. Match by topic, not number. FAQ (FAQPage schema, 5 Q&A): Q1: What is the difference between speed and velocity? A: speed is distance over time - always positive; velocity is displacement over time and carries direction, so it can be zero or negative even after a long trip. Q2: What are the three equations of motion? A: v = u + at, s = ut + (1/2)at2, and v2 = u2 + 2as - valid only for uniform acceleration. Q3: What does the slope of a velocity-time graph give? A: acceleration; the area under the graph gives displacement. Q4: What is free fall? A: motion under gravity alone, with constant acceleration g = 9.8 m/s2 downward - the equations of motion apply with a = g. Q5: Are these the NCERT exercise questions? A: No. EZT pages are original practice on this chapter - use them to prepare for the textbook exercises, not as a substitute for your book. Internal links: /class-11-physics-chapter-1-units-and-measurements (live), /class-12-physics-chapter-1-electric-charges-and-fields (live), /tuitions, /home-tuition-fees-delhi-ncr Schema: LearningResource + FAQPage, same component pattern as existing solutions pages Mobile: responsive
Check: The substitution and final condition match the worked statement.
Watch for this: Do not mix distance with displacement, or speed with velocity; direction and sign decide the answer.
Now try it yourself
Create one new example with different numbers and solve it by the same method.
Check your answer
Keep the final result in your notebook; the page-level Try It prompts are intentionally answer-free.
The worked answer above shows the method; independent practice should not become another answer key.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- u = 5 m/s, a = 3 m/s2, t = 4 s - find v and s.
- A train stops from 30 m/s in 10 s - find acceleration and stopping distance.
- What does the area under a velocity-time graph give?
- A ball dropped from rest hits the ground at 19.6 m/s - from what height was it dropped?
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Questions about this practice page
- What is the difference between speed and velocity?
- speed is distance over time - always positive; velocity is displacement over time and carries direction, so it can be zero or negative even after a long trip.
- What are the three equations of motion?
- v = u + at, s = ut + (1/2)at2, and v2 = u2 + 2as - valid only for uniform acceleration.
- What does the slope of a velocity-time graph give?
- acceleration; the area under the graph gives displacement.
- What is free fall?
- motion under gravity alone, with constant acceleration g = 9.8 m/s2 downward - the equations of motion apply with a = g.
- Are these the NCERT exercise questions for Motion in a Straight Line?
- No. EZT pages are original practice on this chapter - use them to prepare for the textbook exercises, not as a substitute for your book.