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Graphical Representation of Motion - UNSOLVED PRACTICE SET

Class 9

Chapter: Motion | Topic: Graphical Representation of Motion

Study Material.
Class 9

GRAPHICAL REPRESENTATION OF MOTION - UNSOLVED PRACTICE SET

Topic: Graphical Representation of Motion

Time: 40 mins | Marks: 30 | Difficulty: Medium

Multiple Choice Questions

Q1. The slope of a distance–time graph represents:

  1. Acceleration
  2. Displacement
  3. Speed
  4. Force

Q2. The area under a velocity–time graph represents:

  1. Acceleration
  2. Speed
  3. Force
  4. Displacement (distance)

Q3. A horizontal straight line on a velocity–time graph means the object is:

  1. At rest
  2. Moving with uniform acceleration
  3. Moving with constant velocity
  4. Decelerating

Q4. A straight line through the origin on a distance–time graph indicates:

  1. Uniform acceleration
  2. Non-uniform motion
  3. Uniform speed
  4. Rest

Q5. Which graph represents an object that is stationary?

  1. A straight line with positive slope on a d-t graph
  2. A horizontal line on a d-t graph
  3. A parabola on a d-t graph
  4. A straight line with negative slope on a v-t graph

Q6. The slope of a velocity–time graph gives:

  1. Speed
  2. Displacement
  3. Acceleration
  4. Distance

Short Answer Questions

Q7. What does the slope of a distance–time graph represent? How would you calculate the speed of an object from its d-t graph?

Q8. Describe (or sketch) the distance–time graph for:
(a) an object at rest,
(b) an object in uniform motion,
(c) an object in non-uniform (accelerating) motion. What is the key visual difference among these three?

Q9. Describe (or sketch) the velocity–time graph for:
(a) uniform motion,
 (b) uniform acceleration from rest,
(c) uniform deceleration to rest.
What does the slope represent in each case?

Q10. How do you find the distance covered by an object from its velocity–time graph? Explain using the example of an object moving at 15 m/s for 8 seconds.

Q11. A velocity–time graph shows a straight line starting at v = 0 at t = 0 and reaching v = 30 m/s at t = 10 s.
(a) What is the acceleration?
(b) What is the distance covered in 10 s using the graph?
(c) Is this uniform or non-uniform acceleration?

Q12. Explain what a 'parabolic' distance–time graph means physically. Why does a freely falling object produce a curved (parabolic) d-t graph rather than a straight line?

Long Answer Questions

Q13. A velocity–time graph of a vehicle shows the following: • 0 to 5 s: velocity increases from 0 to 25 m/s (uniform acceleration) • 5 s to 15 s: velocity stays at 25 m/s (constant) • 15 s to 20 s: velocity drops from 25 m/s to 0 (uniform deceleration) 
(a) Describe and sketch this v-t graph.
(b) Calculate the acceleration in phase 1.
(c) Calculate the deceleration in phase 3.
(d) Calculate the distance covered in each phase using the area under the graph.
(e) What is the total distance for the full 20 s?

Q14. Use the velocity–time graph method to derive the third equation of motion: v² = u² + 2as.
(a) Draw (or describe) a v-t graph for an object with initial velocity u, final velocity v, acceleration a, and time t.
(b) Show that the area under the graph equals the displacement s = ½(u + v) × t.
(c) Eliminate t using the first equation v = u + at.
(d) Simplify to arrive at v² = u² + 2as.

Q15. Study the following velocity–time data for an auto-rickshaw in Mumbai:    Time (s):    0   4   8   12   16  20   Velocity (m/s): 0  8  16   16   10   0 
(a) Plot (or describe) the v-t graph.
(b) Identify phases of: uniform acceleration, constant velocity, and deceleration.
(c) Calculate the acceleration in each phase.
(d) Calculate the total distance travelled using the areas under each phase.
(e) Is the motion overall uniform or non-uniform?

Numerical / Application-Based Problems

Q16. A distance–time graph has points: (0 s, 0 m), (2 s, 10 m), (4 s, 20 m), (6 s, 20 m), (8 s, 36 m).
(a) Calculate the speed between 0–4 s.
(b) What happened between 4–6 s?
(c) Calculate the speed between 6–8 s.
(d) Is the overall motion uniform or non-uniform? Justify with data.

Q17. A velocity–time graph of a stone thrown upward (g = 10 m/s²) shows a straight line from v = +30 m/s at t = 0 to v = −30 m/s at t = 6 s, passing through v = 0 at t = 3 s.
(a) What is the acceleration (slope)?
(b) What is the physical meaning of v = 0 at t = 3 s?
(c) Calculate the total distance using areas (up and down separately).
(d) What is the total displacement?

Q18. Two cyclists start from the same point. Cyclist X has a v-t graph that is a horizontal line at 10 m/s. Cyclist Y has a v-t graph that is a straight line rising from 0 m/s to 20 m/s in 10 s.
(a) What type of motion is each cyclist in?
(b) Who travels farther in the first 10 seconds? Show by calculating area.
(c) At what time do they have the same speed?
(d) At what time do they have the same distance?
(Note: they will cross — find when.)


Total: 30 Marks | Time: 40 mins

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