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Correct option is C)

$x=16t−2t_{2}$

Now,

Displacement at

$t=0,x=0$

$t=4,x=32$

$t=8,x=0$

$∴$ Distance $=$ (distance from $x=0$ to $x=32$) + (distance from $x=32$ to $x=0$)

$∴$ Distance $=32+32=64m$

Comparing with second equation of motion:

$s=ut+21 at_{2}$

It can be concluded that:

$u=16m/s,a=4m/s_{2}$

Now, using first equation of motion:

$v=u+at$

$v=16−4t$

For $v=0$ we get, $t=4s$

For $v=0$ we get, $t=4s$

Here, direction of velocity will reverse. So total distance travelled will be the sum of displacement in first four seconds and displacement in next four seconds.

Now,

Displacement at

$t=0,x=0$

$t=4,x=32$

$t=8,x=0$

$∴$ Distance $=$ (distance from $x=0$ to $x=32$) + (distance from $x=32$ to $x=0$)

$∴$ Distance $=32+32=64m$

Solve any question of Motion in a Straight Line with:-

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