The displacement of a particle is given by $x = (t - 2)^{2}$ where $x$ is in metres and $t$ in seconds The distance covered by the particles in first $4$ seconds is:

4 m

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8 m

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12 m

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16 m

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Solution

The correct option is B $8 m$
Given $x = {(t - 2)}^{2} = t^{2} - 4 t + 4$
Now at $t = 0$
$x = 4$ m
at $t = 2$ s
$x = 4 - 8 + 4 = 0$ m
and at $t = 4$ s
$x = 16 - 16 + 4 = 4$ m
Thus total distance covered at $t = 4 s = 8$ units.

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

The correct option is B 8mGiven x=(t−2)2=t2−4t+4 Now at t=0x=4 mat t=2sx=4−8+4=0 mand at t=4 sx=16−16+4=4 mThus total distance covered at t=4s=8 units.

The displacement of a particle is given by $x = (t - 2)^{2}$ where $x$ is in metres and $t$ in seconds The distance covered by the particles in first $4$ seconds is:

The correct option is B $8 m$
Given $x = {(t - 2)}^{2} = t^{2} - 4 t + 4$
Now at $t = 0$
$x = 4$ m
at $t = 2$ s
$x = 4 - 8 + 4 = 0$ m
and at $t = 4$ s
$x = 16 - 16 + 4 = 4$ m
Thus total distance covered at $t = 4 s = 8$ units.