Question

# The position of a particle along x-axis at time t is given by x=2+t−3t2. The displacement and the distance travelled in the interval t=0 to t=1 are respectively:

A
2,2.16
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B
0,2
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C
2,2
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D
None of these
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Solution

## The correct option is A −2,2.16x=2+t−3t2⟶1diff. w.r.t. timedxdt=1−6t⟶2again differentiate w.r.t. timedx2dt2=−6Hence motion is retarding becaused2xdt2<0Let at t time velocity of particle is zerodxdt=0=1−6t∴t=16at ∴t=16 velocity of particle is zero, then particle moves in back direction at t=1secNow, displacement= final position − initial position=x(1)−x(0)=2+1−3−2=−2mDistance=∣∣∣x(1)−x(16)∣∣∣+∣∣∣x(16)−x(0)∣∣∣=∣∣∣2+1−3−(2+16−336)∣∣∣+∣∣∣2+16−336−2∣∣∣=∣∣∣−2−16+112∣∣∣+∣∣∣2+16−112−2∣∣∣=∣∣∣−2−112∣∣∣+∣∣∣112∣∣∣=2+112+112=2+16=136=2.16m

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