A body of mass 4 ,kg is accelerated by a cons | vedclass

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(C) The distance travelled by a body in the $n^{	ext{th}}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.For the first second $(n=1

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$6$

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(C) The distance travelled by a body in the $n^{	ext{th}}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.For the first second $(n=1)$: $5 = u + \frac{a}{2}(2(1) - 1) \implies 5 = u + \frac{a}{2}$ (Eq. $i$).For the third second $(n=3)$: $2 = u + \frac{a}{2}(2(3) - 1) \implies 2 = u + \frac{5a}{2}$ (Eq. $ii$).Subtracting Eq. $i$ from Eq. $ii$: $(2 - 5) = (u + \frac{5a}{2}) - (u + \frac{a}{2}) \implies -3 = 2a \implies a = -1.5 \,m/s^2$.The negative sign indicates deceleration.The magnitude of the force is $F = |m 	imes a| = 4 \,kg 	imes 1.5 \,m/s^2 = 6 \,N$.

$A$ body of mass $4 \,kg$ is accelerated by a constant force. It travels a distance of $5 \,m$ in the first second and a distance of $2 \,m$ in the third second. The force acting on the body is: (in $\,N$)

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