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

The NCERT text explicitly states, 'Stopping distance is an important factor considered in setting speed limits, for example, in school zones.' The primary reason for lower speed limits in areas like school zones is to allow drivers more time and distance to stop if unexpected situations (like a child running onto the road) arise, thereby enhancing safety.

The process described involves two phases: first, the time taken to react (reaction time) during which the vehicle continues to move (reaction distance), and second, the time taken for the vehicle to stop after the brakes are applied (braking distance or stopping distance). The NCERT mentions 'Reaction time is the time a person takes to observe, think and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before he slams the brakes of the car is the reaction time.' This implies motion during reaction time. 'stopping distance...is an important factor for road safety and depends on the initial velocity (v0) and the braking capacity'.

From the formula $d_s = -v_0^2 / (2a)$, we can see that stopping distance ($d_s$) is inversely proportional to the magnitude of deceleration ($|a|$). A larger deceleration means the vehicle can slow down and stop more quickly, hence covering a shorter distance. The negative sign for 'a' implies deceleration, so we consider its magnitude.

As per the NCERT example, the reaction time ($t_r$) is calculated using the formula $d = \frac{1}{2} g t_r^2$. Rearranging for $t_r$: $t_r = \sqrt{\frac{2d}{g}}$. Given $d = 21.0 \text{ cm} = 0.21 \text{ m}$ and $g = 9.8 \text{ m/s}^2$. So, $t_r = \sqrt{\frac{2 \times 0.21}{9.8}} = \sqrt{\frac{0.42}{9.8}} = \sqrt{0.0428...} \approx 0.207 \text{ s}$. Rounding to two significant figures, $t_r \approx 0.21 \text{ s}$.

The NCERT definition of reaction time states: 'Reaction time is the time a person takes to observe, think and act.' It also mentions that 'Reaction time depends on complexity of the situation and on an individual,' disproving the first two options. The third option defines only the 'act' component, not the full process.

According to 'POINTS TO PONDER' point 2, 'If a particle is speeding up, acceleration is in the direction of velocity; if its speed is decreasing, acceleration is in the direction opposite to that of the velocity.' This directly answers the question.

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.'