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Question

# The position of a particle along x-axis at time t is given by x=1+t−t2 m. The distance travelled by the particle in first 2 sec is

A
1 m
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B
2 m
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C
2.5 m
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D
3 m
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Solution

## The correct option is C 2.5 mx=1+t−t2 At t=0 sec, x0=1 m We know that, v=dxdt=1−2t So, at t=12 sec,v=0 And displacement , x12=1+12−14=54=1.25 m Distance travelled by the particle from t=0 to t=12 sec =|x12−x0|=1.25 m−1 m=0.25 m Position of particle at the end of 2 sec is given by x2=1+2−22=−1 m Distance travelled by the particle t=12 sec to t=2 sec =|x2−x12|=|−1m−1.25 m|=2.25 m Total distance travelled by the particle from t=0 sec to t=2 sec will be ∴x=0.25 m+2.25 m=2.5 m

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