Question

The displacement $ x $ versus time graph is shown below.
 
The displacement $ x $ is plotted against time $ t $. Choose the correct answer from the options given below:

• A. The average velocity during 0 to 3 s is 10 m/s
• B. The average velocity during 3 to 5 s is 0 m/s
• C. The instantaneous velocity at \( t = 2 \) s is 5 m/s
• D. The average velocity during 5 to 7 s is the same as instantaneous velocity at \( t = 6.5 \) s

Hint:

The average velocity is the total displacement divided by the total time, while instantaneous velocity is the slope of the displacement-time graph at a given point.

Answer 1

The Correct Option is D

To solve this problem, we need to analyze the displacement versus time graph given and understand the definitions of average velocity and instantaneous velocity.

Definitions

• Average Velocity: It is defined as the total displacement divided by the total time taken. Mathematically, \(v_{\text{avg}} = \frac{\Delta x}{\Delta t}\).
• Instantaneous Velocity: It is the velocity of an object at a specific instant of time. It is obtained by finding the slope of the tangent to the graph at that point.

Analysis of Options

1. The average velocity during 0 to 3 s is 10 m/s:
   In this interval, the object moves from point A (at 0 s and \(x = -5 \, \text{m}\)) to point C (at 3 s and \(x = 5 \, \text{m}\)).
   Total displacement = \(5 - (-5) = 10 \, \text{m}\).
   Average velocity = \(\frac{10 \, \text{m}}{3 \, \text{s}} = \frac{10}{3} \, \text{m/s} \neq 10 \, \text{m/s}\).
   This option is incorrect.
2. The average velocity during 3 to 5 s is 0 m/s:
   From 3 s to 5 s, the object moves from point C (5 m) to point D (5 m), so there is no displacement.
   Total displacement = \(5 - 5 = 0 \, \text{m}\).
   Average velocity = \(\frac{0}{2} = 0 \, \text{m/s}\).
   This option is correct, but let's check the other options.
3. The instantaneous velocity at \( t = 2 \) s is 5 m/s:
   At \(t = 2 \, \text{s}\), the graph shows a linear segment from A to C with a slope of \(5 \, \text{m/s}\).
   This option is consistent, but not the only correct one.
4. The average velocity during 5 to 7 s is the same as instantaneous velocity at \( t = 6.5 \) s:
   From 5 s to 7 s, the object moves from point E (10 m at 5 s) to point G (0 m at 7 s).
   Total displacement = \(0 - 10 = -10 \, \text{m}\).
   Average velocity = \(\frac{-10}{2} = -5 \, \text{m/s}\).
   At \( t = 6.5 \) s, the graph section EF has a slope of -5 m/s.
   This option is correct and matches the provided correct answer.

Conclusion

The correct answer is: The average velocity during 5 to 7 s is the same as instantaneous velocity at \( t = 6.5 \) s.

Answer 2

To find the average and instantaneous velocities, we use the formulas: 
1. Average Velocity: \[ \bar{v} = \frac{\Delta x}{\Delta t} \] where \( \Delta x \) is the change in displacement and \( \Delta t \) is the time interval. 
2. Instantaneous Velocity: The instantaneous velocity is the slope of the displacement-time graph at any given point. 
Now, let's calculate each part: 
- (A) Average velocity during 0 to 3 s: The displacement changes from \( 0 \) to \( 30 \, \text{m} \) in 3 seconds: \[ \bar{v} = \frac{30 - 0}{3 - 0} = \frac{30}{3} = 10 \, \text{m/s} \] 
Hence, (A) is correct. 
- (B) Average velocity during 3 to 5 s: The displacement does not change from \( t = 3 \) to \( t = 5 \) s, so: \[ \bar{v} = \frac{0 - 0}{5 - 3} = 0 \, \text{m/s} \] Hence, (B) is correct. 
- (C) Instantaneous velocity at \( t = 2 \) s: From the graph, the slope of the line at \( t = 2 \) s is: \[ v = 5 \, \text{m/s} \] Hence, (C) is correct. 
- (D) Average velocity during 5 to 7 s: The displacement changes from \( 10 \, \text{m} \) to \( 10 \, \text{m} \) (no change) in 2 seconds, so: \[ \bar{v} = \frac{10 - 10}{7 - 5} = 0 \, \text{m/s} \] The instantaneous velocity at \( t = 6.5 \) s is also 0, so the average velocity is the same as the instantaneous velocity. Hence, (D) is correct. 
- (E) Average velocity from \( t = 0 \) to \( t = 9 \) s: The displacement returns to zero at \( t = 9 \) s, so the average velocity is: \[ \bar{v} = \frac{0 - 0}{9 - 0} = 0 \, \text{m/s} \] Hence, (E) is correct. 
Thus, the correct answer is (4).