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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Solution

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

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A body with an initial velocity of 5m/s has an acceleration of 2 m/s2 . What is the distance covered by it in the fourth second? __

Let the distance covered after 3s be S3 . Then we have S3 = (5 x 3) + (0.5 x 2 x 32) = 15 + 9 = 24m Let the distance covered after 4s beS4 . S4 = (5 x 4) + (0.5 x 2 x 42) = 20 + 16 = 36 m Therefore Distance covered in 4th second = S4 - S3 =( 36 -24)m = 12m

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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