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Question

The position of a particle along x-axis at time t is given by x=2+t3t2. 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 2,2.16
x=2+t3t21
diff. w.r.t. time
dxdt=16t2
again differentiate w.r.t. time
dx2dt2=6
Hence motion is retarding because
d2xdt2<0
Let at t time velocity of particle is zero
dxdt=0=16t
t=16
at t=16 velocity of particle is zero, then particle moves in back direction at t=1sec
Now, displacement= final position initial position
=x(1)x(0)
=2+132
=2m
Distance=x(1)x(16)+x(16)x(0)
=2+13(2+16336)+2+163362
=216+112+2+161122
=2112+112=2+112+112
=2+16=136=2.16m

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