Motion in a Straight Line MCQs for NEET — Physics Questions with Answers

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).

The text states, 'Since velocity is a quantity having both magnitude and direction, a change in velocity may involve either or both of these factors. Acceleration, therefore, may result from a change in speed (magnitude), a change in direction or changes in both.'

Figure 2.4 (c) shows a position-time graph as a straight line, and it is explicitly labeled as 'zero acceleration'. The text also states, 'it is a straight line for zero acceleration' in the context of position-time graphs.

According to the provided text and the derived expression in Example 2.6, the stopping distance is given by $d_s = \frac{-v_0^2}{2a}$. This shows that the stopping distance is proportional to the square of the initial velocity ($v_0^2$). The context explicitly states, 'Thus, the stopping distance is proportional to the square of the initial velocity.'

The context states, 'Doubling the initial velocity increases the stopping distance by a factor of 4 (for the same deceleration).' This is because stopping distance ($d_s$) is proportional to the square of the initial velocity ($v_0^2$). If $v_0$ becomes $2v_0$, then $d_s'$ will be proportional to $(2v_0)^2 = 4v_0^2$, making the new stopping distance $4x$.