CBSE Class 11 Physics NCERT Solutions: Motion in a Straight Line

The average velocity over a time interval is defined as the net displacement divided by the time interval. For the average velocity to vanish, the net displacement must be zero. This means the object must return to its starting position within the time interval (0, T).

Let's analyze the given graphs:

• Graph (a): This graph shows position increasing linearly with time. The displacement will always be positive, so the average velocity cannot be zero.
• Graph (b): This graph shows that for a single value of position (x), there can be two different values of time. If we choose a time T such that the position at T is the same as the position at time 0 (which is usually assumed to be the origin or a reference point), then the displacement is zero, and hence the average velocity is zero. This type of graph represents motion where the object might move forward, stop, and then move backward, or vice versa, potentially returning to its initial position.
• Graph (c): This graph shows position increasing and then decreasing, but it does not necessarily return to the initial position.
• Graph (d): This graph shows position decreasing linearly with time. The displacement will always be negative, so the average velocity cannot be zero.

Therefore, graph (b) is the one where the average velocity over the time interval (0, T) can vanish for a suitably chosen T, as it allows for the possibility of returning to the initial position.

We are given that the ground floor is the origin, and the upward direction is positive. The lift is moving downwards from the 8th floor towards the 4th floor.

• Position (x): Since the lift is between the 8th and 4th floors, and the ground floor (origin) is below it, its position is below the origin. Therefore, the position $x$ is negative ().
• Velocity (v): The lift is moving downwards. Since the upward direction is positive, the downward direction is negative. Thus, the velocity $v$ is negative ().
• Acceleration (a): The lift is coming from the 8th floor and is about to reach the 4th floor. This implies that the lift is slowing down (retarding) as it approaches the 4th floor. Retarding motion means the acceleration is in the opposite direction to the velocity. Since the velocity is downwards (negative), the acceleration must be upwards (positive). Therefore, the acceleration $a$ is positive ($a\; >\; 0$).

Combining these, we have , , and $a\; >\; 0$.

This corresponds to option (a).

The condition given is that the instantaneous speed $v$ satisfies . This means the magnitude of the velocity is always less than $v\_0$. The velocity can be positive or negative, but its magnitude is bounded.

• Option (a): The displacement can be negative if the object moves in the negative direction. For example, if an object starts at $x=0$ and moves with a constant velocity of $-v\_0/2$ for time $T$, its displacement will be $-v\_0\; T\; /\; 2$, which is negative. So, (a) is incorrect.
• Option (b): The maximum possible displacement in the positive direction in time $T$ occurs if the velocity is constantly $+v\_0$ (or just below it). This maximum displacement would be $v\_0\; T$. Similarly, the maximum possible displacement in the negative direction occurs if the velocity is constantly $-v\_0$ (or just below it in magnitude). This minimum displacement would be $-v\_0\; T$. Since the actual speed $v$ is always less than $v\_0$, the displacement $x$ in time $T$ must lie between $-v\_0\; T$ and $+v\_0\; T$. Thus, . This statement is correct.
• Option (c): The acceleration is not necessarily non-negative. For example, if an object moves with a constant velocity $v\; =\; -v\_0/2$, its acceleration is zero. If it moves with $v\; =\; -v\_0/2$ for some time and then $v\; =\; +v\_0/2$, the acceleration could be positive or negative during these intervals. The speed constraint does not restrict the sign of acceleration. So, (c) is incorrect.
• Option (d): Turning points occur when the velocity changes sign. Since the velocity can be positive or negative (as long as its magnitude is less than $v\_0$), the motion can have turning points. For instance, if an object moves in the positive direction and then reverses to move in the negative direction, it has a turning point. So, (d) is incorrect.

Therefore, the correct statement is (b).

The average speed is defined as the total distance traveled divided by the total time taken.

Let the total distance be $l$. The distance is divided into two equal halves, so each half is $\backslash frac\{l\}\{2\}$.

• Time taken to travel the first half distance:

  Distance = $\backslash frac\{l\}\{2\}$

  Speed = $v\_1$

  Time $t\_1\; =\; \backslash frac\{\backslash text\{Distance\}\}\{\backslash text\{Speed\}\}\; =\; \backslash frac\{l/2\}\{v\_1\}\; =\; \backslash frac\{l\}\{2v\_1\}$

• Time taken to travel the second half distance:

  Distance = $\backslash frac\{l\}\{2\}$

  Speed = $v\_2$

  Time $t\_2\; =\; \backslash frac\{\backslash text\{Distance\}\}\{\backslash text\{Speed\}\}\; =\; \backslash frac\{l/2\}\{v\_2\}\; =\; \backslash frac\{l\}\{2v\_2\}$

• Total time taken:

  $T\; =\; t\_1\; +\; t\_2\; =\; \backslash frac\{l\}\{2v\_1\}\; +\; \backslash frac\{l\}\{2v\_2\}$

  $T\; =\; \backslash frac\{l\}\{2\}\; \backslash left(\; \backslash frac\{1\}\{v\_1\}\; +\; \backslash frac\{1\}\{v\_2\}\; \backslash right)$

  $T\; =\; \backslash frac\{l\}\{2\}\; \backslash left(\; \backslash frac\{v\_2\; +\; v\_1\}\{v\_1v\_2\}\; \backslash right)$

• Average speed:

  $\backslash text\{Average\; Speed\}\; =\; \backslash frac\{\backslash text\{Total\; Distance\}\}\{\backslash text\{Total\; Time\}\}\; =\; \backslash frac\{l\}\{T\}$

  $\backslash text\{Average\; Speed\}\; =\; \backslash frac\{l\}\{\backslash frac\{l\}\{2\}\; \backslash left(\; \backslash frac\{v\_1\; +\; v\_2\}\{v\_1v\_2\}\; \backslash right)\}$

  $\backslash text\{Average\; Speed\}\; =\; \backslash frac\{1\}\{\backslash frac\{1\}\{2\}\; \backslash left(\; \backslash frac\{v\_1\; +\; v\_2\}\{v\_1v\_2\}\; \backslash right)\}$

  $\backslash text\{Average\; Speed\}\; =\; \backslash frac\{2\; v\_1v\_2\}\{v\_1\; +\; v\_2\}$

This is the harmonic mean of the two speeds.

Therefore, the correct option is (c).