- Dr.Sanjaykumar Pawar
NCERT Solutions
Class 11 Physics
Chapter 2: Motion in a Straight Line
Exercise 2.15 to 2.18
Question 2.15
Suggest a suitable physical situation for each of the following graphs (Fig. 2.12).
(a) Position-Time Graph (x–t)
Observation:
- The particle remains at rest initially.
- Then moves away from the origin with constant velocity.
- Returns towards the origin with constant velocity.
- Finally comes to rest again.
Suitable Physical Situation:
A person standing on a straight road starts walking uniformly away from the origin, reaches a point A, turns back, crosses the origin at B and finally stops at another point.
(b) Velocity-Time Graph (v–t)
Observation:
- Velocity decreases linearly.
- Changes sign repeatedly.
- Magnitude decreases after every interval.
Suitable Physical Situation:
A ball thrown vertically upward repeatedly strikes the ground and rebounds. Each collision reduces its speed due to loss of energy.
(c) Acceleration-Time Graph (a–t)
Observation:
- Acceleration is zero for most of the time.
- A sudden positive acceleration occurs for a short interval.
Suitable Physical Situation:
A vehicle moving uniformly receives a sudden push or the accelerator is pressed for a short duration.
Question 2.16
Figure 2.13 gives the x–t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position (x), velocity (v) and acceleration (a) at: (i) t = 0.3 s (ii) t = 1.2 s (iii) t = –1.2 s
Principle Used:
(i) At t = 0.3 s
- x is negative.
- Slope is negative.
- Acceleration is positive.
(ii) At t = 1.2 s
- x is positive.
- Slope is positive.
- Acceleration is negative.
(iii) At t = –1.2 s
- x is positive.
- Slope is positive.
- Acceleration is negative.
| Time | Position (x) | Velocity (v) | Acceleration (a) |
|---|---|---|---|
| 0.3 s | Negative | Negative | Positive |
| 1.2 s | Positive | Positive | Negative |
| –1.2 s | Positive | Positive | Negative |
Question 2.17
Figure 2.14 gives the x–t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest and in which is it least? Give the sign of average velocity for each interval.
Concept:
Since all intervals have equal duration, the interval with the greatest change in position will have the greatest average speed.
Interval 1
- Position increases slowly.
- Average velocity is positive.
Interval 2
- Position increases rapidly.
- Average velocity is positive.
Interval 3
- Position decreases rapidly.
- Average velocity is negative.
Least Average Speed = Interval 1
| Interval | Sign of Average Velocity |
|---|---|
| 1 | Positive |
| 2 | Positive |
| 3 | Negative |
Question 2.18
Figure 2.15 gives a speed-time graph of a particle moving in a constant direction. Three equal intervals of time are shown. (a) In which interval is the average acceleration greatest in magnitude? (b) In which interval is the average speed greatest? (c) Choosing the positive direction as direction of motion, give signs of velocity and acceleration in the three intervals. (d) What are the accelerations at points A, B, C and D?
(a) Average Acceleration
Since all intervals have equal duration, the third interval shows the largest change in speed.
(b) Average Speed
The average speed is greatest in the interval where the speed values are highest.
(c) Signs of Velocity and Acceleration
| Interval | Velocity | Acceleration |
|---|---|---|
| 1 | Positive | Positive |
| 2 | Positive | Negative |
| 3 | Positive | Positive |
(d) Accelerations at A, B, C and D
Acceleration is equal to the slope of the speed-time graph.
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| NCERT Class 11 Physics Chapter 2 Motion in a Straight Line Exercise 2.15 to 2.18 solved with graphs, SHM concepts, velocity and acceleration analysis. |
