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The displacement of a particle is given by x = (t – 2)2 where x is in meters and t in seconds. The distance covered by the particle in the first 4 seconds is:

  • a)
    4 m

  • b)
    12 m

  • c)
    16 m

  • d)
     8 m

Correct answer is option 'D'. Can you explain this answer?
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The displacement of a particle is given by x = (t – 2)2where x i...
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The displacement of a particle is given by x = (t – 2)2where x i...
X = (t - 2)^2 --------(1)

dX/dt =u=2(t - 2) = 2t-4

i.e, at t = 2 sec , u= 0 m/s
it means particle comes to the momentarly rest at t= 2 sec.

Now, a = du/dt= 2 m/s^2

at t=0 , x = 4 m and at t= 2sec (when 1st u=0) x = 0m ( particle move from 4 m to 0m towards -ve x-axis)
i. e, in 1st 2 sec it travels 4m ............. (2)
but at t= 2 sec , v= 0m/s and a= 2 m/s^2
so,for next 2 sec
X = 1/2 (at^2)
= 1/2× 2× (2) ^2
=4m ........................ (3)

therefore, total distance covered in 4 sec = 8m (equ 2+ equ 3)
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The displacement of a particle is given by x = (t – 2)2where x is inmeters and t in seconds. The distance covered by the particle in the first 4 seconds is:a)4 mb)12 mc)16 md)8 mCorrect answer is option 'D'. Can you explain this answer?
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