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  4. If the rate of change in velocity w.r.t time is constant and its position after 6th second will be same as that after 11th second then the particle returns to the starting point at time t equals to

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Q. If the rate of change in velocity w.r.t time is constant and its position after $6^{th}$ second will be same as that after $11^{th}$ second then the particle returns to the starting point at time t equals to

Solution:

$T = t_1 +t_2 $
$= 17\,s$
image
Here acceleration is constant and body returns, therefore initial velocity and acceleration must be in opposite direction as shown above. Therefore particle returns after $17$ seconds.