A particle having initial velocity u moves with a constant acceleration a for a time $t .$ Find the displacement of the particle in the last $1$ second.

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Solution

The distance travelled in the $n^{t h}$ second is given as,

$s_{n} = u + \frac{1}{2} a (2 n - 1)$

Thus, the distance travelled in $t^{t h}$ second is,

$s_{n} = u + \frac{1}{2} a (2 t - 1)$

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