The position of a particle vary x=4t−2t2. The distance covered by particle in 4 s is

[d] Idea In expressions x=4t−2t2 the ′x′ represents displacement not distance.$x=4t-2t2,v=4\left(1-t\right)\text{and}att=1,v=0$$Thevelocityofthepartic\le becomezeroatt=1s\text{and}thenitmoves\to wardsthe\neg ativedirection.$⇒            x1 for 1st second =4−2×1=+2m$\Rightarrow x2f\text{or}2nd\sec ond$            =4×2−2×4=8−8=0$So,firstpartic\le hascovers2m\text{and}thenitcomesback2m.$So, the total distance covered =2+2=4 m.\displaystyle{T}{E}{S}{T}{E}{d}\ge{T}{h}{e}{r}{e}{c}{o}\underline{{d}}{b}{e}{d}\Leftrightarrow{e}{r}{e}{n}{t}{r}{e}{l}{a}{t}{i}{o}{n}{s}{p}{o}{s}{s}{i}{b}\le{i}.{e}.,{v}={2}{x}+{5}{\quad\text{and}\quad}{o}\ne{c}{a}{n}{a}{s}{k}{t}{h}{e}{r}{e}{l}{a}{t}{i}{o}{n}{b}{e}{t}{w}{e}{e}{n}{x}{\quad\text{and}\quad}{t}{\mathfrak{{o}}}{m}{t}{h}{e}{g}{i}{v}{e}{n}{e}{q}{u}{a}{t}{i}{o}{n}.{H}{e}{r}{e},{p}{u}{t}{v}={\left.{d}{x}\right.}{\left.{d}{t}\right.}{\quad\text{and}\quad}\int{e}{g}{r}{a}{t}{e}{t}{h}{e}{\exp{{r}}}{e}{s}{s}{i}{o}{n}\to\ge{t}{t}{h}{e}{r}{e}{l}{a}{t}{i}{o}{n}{b}{e}{t}{w}{e}{e}{n}{x}{\quad\text{and}\quad}{t}.