Practice free Motion in a Straight Line (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
Reaction time is best described as the:
Example 2.7 clearly defines reaction time: 'Reaction time is the time a person takes to observe, think and act.' It encompasses the entire cognitive and motor process from stimulus recognition to initial response.
In a reaction time experiment where a ruler (under free fall) travels a distance $d$ before being caught, the relationship between $d$ and the reaction time $t_r$ is given by:
From Example 2.7, an object under free fall (like the ruler) starts with an initial velocity $v_0 = 0$. The distance travelled ($d$) is related to time ($t_r$) by the kinematic equation $d = v_0 t_r + \frac{1}{2} a t_r^2$. Since $v_0 = 0$ and acceleration $a = g$, the equation simplifies to $d = \frac{1}{2} g t_r^2$. Note: The solution in the text uses $a = -g$ when defining the direction of motion relative to the coordinate system, but for the magnitude of distance traveled downwards, 'g' is used positively.
A driver sees an obstacle and takes $0.5 \text{ s}$ to react before applying the brakes. During this reaction time, the car is moving at a constant speed of $20 \text{ m/s}$. How far does the car travel during this reaction time?
During the reaction time, the car continues to move at its initial speed before deceleration begins. The distance covered during reaction time can be calculated as distance = speed × time. So, distance = $20 \text{ m/s} \times 0.5 \text{ s} = 10 \text{ m}$.
Which of the following factors is explicitly mentioned as influencing an individual's reaction time?
Example 2.7 states, 'Reaction time depends on complexity of the situation and on an individual.' Therefore, the complexity of the situation directly influences reaction time.
If the deceleration ('a') applied by the brakes on a vehicle were to increase (meaning the braking force is stronger), what would be the effect on the stopping distance, assuming the initial velocity remains constant?
The stopping distance formula is $d_s = \frac{-v_0^2}{2a}$. Here, '$a$' represents deceleration, which is inherently negative. If we consider the magnitude of deceleration, a larger magnitude of 'a' (meaning stronger braking) would lead to a smaller stopping distance because 'a' is in the denominator. The faster you can decelerate, the shorter the distance needed to stop.
Why is 'stopping distance' considered an important factor in setting speed limits, especially in school zones?
The text explicitly states, 'Stopping distance is an important factor considered in setting speed limits, for example, in school zones.' This is because school zones have a higher risk of unexpected pedestrian (children) presence, necessitating shorter stopping distances at lower speeds to prevent accidents.
A vehicle travels half the distance with speed $\vartheta$ and the remaining distance with speed $2\vartheta$. Its average speed is:
$\bar v = \dfrac{d}{d/(2\vartheta) + d/(4\vartheta)} = \dfrac{d}{3d/(4\vartheta)} = \dfrac{4\vartheta}{3}$.
A bullet from a gun is fired on a rectangular wooden block with velocity $u$. When bullet travels $24\ \text{cm}$ through the block along its length horizontally, velocity of bullet becomes $\dfrac{u}{3}$. Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is:
$v^2 - u^2 = 2as$. From $u$ to $u/3$ in 24 cm: $\dfrac{8u^2}{9} = 2a(24)$. From $u/3$ to 0: $\dfrac{u^2}{9} = 2a\,s\Rightarrow s = 3$ cm. Total $= 27$ cm.
A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity $4\ \text{m s}^{-1}$. The ball strikes the water surface after $4\ \text{s}$. The height of bridge above water surface is (Take $g = 10\ \text{m s}^{-2}$):
Taking up positive from bridge: $-h = ut - \tfrac{1}{2}gt^2 = 4(4) - \tfrac{1}{2}(10)(16) = 16 - 80 = -64$, so $h = 64$ m.
The velocity ($v$) – time ($t$) plot of the motion of a body is shown below:
The acceleration ($a$) – time ($t$) graph that best suits this motion is:
Trapezoidal $v$–$t$ → positive constant $a$, then $a = 0$, then negative constant $a$.
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