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

Boat covers distance of 16km in a still water in 2 hours.

i.e. $v_B=\frac{16}{2}=8km/hr$

Now velocity of water $\Rightarrow v_w=4km/hr$.

Time taken for going upstream

$t_1=\frac{8}{v_B-v_w}=\frac{8}{8-4}=2hr$

(As water current oppose the motion of boat)

Time taken for going down stream

$t_2=\frac{8}{v_B+v_w}=\frac{8}{8+4}=\frac{8}{12}hr$

(As water current helps the motion of boat)

∴ Total time $=t_1+t_2=\left(2+\frac{8}{12}\right)hr$ or 2hr 40min

Relative velocity $=10+5=15m/s$.

Time taken by the bird to cross the train $=\frac{120}{15}=8 \sec$

The relative velocity $v_{rel.}=60-(-30)=90km/hr.$

Distance between the train $s_{rel.}=90km,$

∴ Time when they collide $=\frac{s_{rel.}}{v_{rel.}}=\frac{90}{90}=1hr.$ 

$v_T=\sqrt{v_{TC}^2+v_C^2}$ = $\sqrt{(25{\sqrt{3)}}^2+{\left(25\right)}^2}$

= $\sqrt{1875+625}$ = $\sqrt{2500}$ = 50 km/hr

Average acceleration $=\frac{\mathrm{Change} \mathrm{in} \mathrm{velocity}}{\mathrm{Total} \mathrm{time}}$

                  $$\begin{gathered} \mathrm{a}=\frac{|{\mathrm{v}}_{\mathrm{f}}-{\mathrm{v}}_{\mathrm{i}}|}{∆\mathrm{t}}=\frac{|40\mathrm{j}-30\mathrm{i}|}{∆\mathrm{t}} \\ =\frac{\sqrt{{40}^2+{30}^2}}{10}=\frac{\sqrt{1600+900}}{10} \\ = 5 {\mathrm{ms}}^{-2} \end{gathered}$$

The NCERT text states that stopping distance depends on initial velocity ($v_0$) and braking capacity (deceleration, $-a$). While reaction time contributes to the total stopping distance (reaction distance + braking distance), the stopping distance specifically refers to the distance traveled after brakes are applied, which is primarily influenced by initial velocity and deceleration. The formula provided, $d_s = -v_0^2 / (2a)$, does not include mass. Although mass affects the deceleration for a given braking force, the question asks about factors directly influencing the stopping distance given a deceleration.

According to the NCERT text, 'Thus, the stopping distance is proportional to the square of the initial velocity. Doubling the initial velocity increases the stopping distance by a factor of 4 (for the same deceleration).' This is derived from the formula $d_s = -v_0^2 / (2a)$. If $v_0$ becomes $2v_0$, then $d_s$ becomes $(2v_0)^2 / (2a) = 4v_0^2 / (2a)$, which is 4 times the original stopping distance.

The NCERT text states, 'Reaction time depends on complexity of the situation and on an individual.' Factors like fatigue, distraction, or intoxication (e.g., alcohol) would impair an individual's ability to observe, think, and act quickly, thereby increasing reaction time. The other options describe conditions that would likely lead to a decreased or normal reaction time.

The NCERT example states, 'The ruler drops under free fall. Therefore, $v_0 = 0$, and $a = -g = -9.8 \text{ m s}^{-2}$. The distance travelled $d$ and the reaction time $t_r$ are related by $d = \frac{1}{2} g t_r^2$.' This is derived from the equation of motion for constant acceleration, $s = ut + \frac{1}{2}at^2$, where initial velocity $u=0$, acceleration $a=g$, and distance $s=d$.

The NCERT text provides data: 'the braking distance was found to be 10 m, 20 m, 34 m and 50 m corresponding to velocities of 11, 15, 20 and 25 m/s'. Therefore, for a velocity of $25 \text{ m/s}$, the braking distance is $50 \text{ m}$ based on the given empirical data.