NEET Practice Questions (MCQs) with Answers & Solutions

As per 'POINTS TO PONDER' point 4, 'The zero velocity of a particle at any instant does not necessarily imply zero acceleration at that instant. A particle may be momentarily at rest and yet have non-zero acceleration. For example, a particle thrown up has zero velocity at its uppermost point but the acceleration at that instant continues to be the acceleration due to gravity.'

The text states, 'Note that the graph curves upward for positive acceleration; downward for negative acceleration and it is a straight line for zero acceleration.' Therefore, a curve upward signifies positive acceleration.

From the text: 'The acceleration at an instant is the slope of the tangent to the v–t curve at that instant.' This directly links instantaneous acceleration to the slope of the velocity-time graph.

From 'POINTS TO PONDER' point 3: 'For example, if the vertically upward direction is chosen to be the positive direction of the axis, the acceleration due to gravity is negative. If a particle is falling under gravity, this acceleration, though negative, results in increase in speed.'

As noted in 'POINTS TO PONDER' point 5: 'The kinematic equations of motion [Eq. (2.9)], the various quantities are algebraic, i.e. they may be positive or negative. The equations are applicable in all situations (for one dimensional motion with constant acceleration) provided the values of different quantities are substituted in the equations with proper signs.'

From the 'MOTION IN A LANE' section: 'Therefore, the direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion.'

The text states: 'We see that as $\Delta t$ decreases, the direction of $\Delta v$ changes and consequently, the direction of the acceleration changes. Finally, in the limit $\Delta t \to 0$ [Fig. 3.15(d)], the average acceleration becomes the instantaneous acceleration and has the direction as shown.' The point is that $\Delta v$ changes direction as $\Delta t$ decreases.

The text clarifies, 'Note that in one dimension, the velocity and the acceleration of an object are always along the same straight line (either in the same direction or in the opposite direction). However, for motion in two or three dimensions, velocity and acceleration vectors may have any angle between 0° and 180° between them.'

From Example 2.1 in the text, the velocity expression is $v = \frac{dx}{dt} = 2bt$. Substituting $b = 2.5 \text{ m s}^{-2}$ and $t = 2.0 \text{ s}$: $v = 2 \times 2.5 \times 2.0 = 10.0 \text{ m s}^{-1}$.

Figure 2.3 (b) is described as 'Motion in positive direction with negative acceleration'. This graph shows a straight line with a negative slope (negative acceleration), meaning velocity decreases over time, but it starts with a positive velocity (moving in the positive direction).