1.

A particle is moving along x-axis whose positionvaried with time t as x 22 -4t + 1, where x isin meter and t is in second. The distance travelledby particle in first 2 second is 、ht-h(1) 2 m(2) 1 m16(4) 8 m

Answer»

given x = 2t²-4t+1

at t = 0 , x = 1 and at = 2 , x = 1

now , maximum displacement is when dx/dt = 0 => 4t-4 = 0 => t = 1 sec and maximum displacement is = 2(1)²-4(1)+1 = 2-4+1 = -1

so, maximum displacement is -1-1 = -2

and in t = 0 to 2 sec , particle goes for 1 to -1 and then comes back at x = 1, (t = 2 sec)

so, net distance = |-2| +|-2| = 4m


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