question_answer

The displacement of a particle is given by \[x\text{ }=\text{ }{{\left( t\text{ }-\text{ }2 \right)}^{2}}\], where x is in meters and t in seconds. The distance covered by the particle in first 4 s is

A) 8m                               

B) 4m

C) 12m                             

D) 16m

Correct Answer: A

Solution :

given, \[x\left( t \right)={{\left( t-2 \right)}^{2}}..(i)\]

Velocity of particle at any time \['t'\]\[=\frac{dx}{dt}\]\[\Rightarrow v\left( t \right)=2\left( t-2 \right)..(ii)\]

Let us find the time at which velocity is zero.

i.e., \[v=0\Rightarrow 2\left( t-2 \right)=0\]or \[t=2s\]

So, before 4sis completed, the particles velocity becomes zero and it takes a turn. Acceleration of particle

\[=\frac{dv}{dt}=2m/{{s}^{2}}..(iii)\]

The motion of the particle along straight line can be seen as

Total distance =path length \[\left( AB+BC \right)\]

For first 2 s;

Using,\[x\left( t \right)={{v}_{0}}t+1/2a{{t}^{2}}\]

\[\because At\,t=0;v\left( 0 \right)={{v}_{0}}=-4m/s\]

\[\left[ \operatorname{put}\,t=0inEq.(ii) \right]\]

Also, \[a=+2m/{{s}^{2}}\left[ \operatorname{from}\,Eq.\left( \operatorname{iii} \right) \right]\]

\[\Rightarrow {{x}_{1}}\left( t \right)=-4\times 2+1/2\times 2\times {{\left( 2 \right)}^{2}}\]

\[=-8+4=-4\]

Distance during this interval\[=\left| x\left( t \right) \right|=4\operatorname{m}\]

For next 2s;\[{{v}_{0}}=v\left( 2 \right)=0m/s\]

\[a=2m/s\]

\[\Rightarrow {{x}_{2}}\left( t \right)=0+1/2\times 2\times {{\left( 2 \right)}^{2}}=4\]

\[\therefore total distance =4+4=8m\]