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;\[{{v}_{0}}=v\left( 2 \right)=0m/s\] |
\[a=2m/s\] |
\[\Rightarrow {{x}_{2}}\left( t \right)=0+1/2\times 2\times {{\left( 2 \right)}^{2}}=4\] |
\[\therefore total distance =4+4=8m\] |
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