Question

A body with an initial velocity of 5m/s has an acceleration of 2 $m/s^2$ . What is the distance covered by it in the fourth second?

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Answer 1

Let the distance covered after 3s be $S_3$ _.

Then we have

$S_3$ = (5 x 3) + (0.5 x 2 x $3^2$)

= 15 + 9 = 24m

Let the distance covered after 4s be$S_4$ _.

$S_4$ = (5 x 4) + (0.5 x 2 x $4^2$)

= 20 + 16 = 36 m

Therefore Distance covered in 4th second = $S_4$ - $S_3$

=( 36 -24)m

= 12m