A particle moving with a uniform acceleration | vedclass

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(A) Let the initial velocity be $u$ and uniform acceleration be $a$.For the first interval of $t = 4 \,sec$,the distance $s_1 = 24 \,m$:$s = ut + \fra

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$1$

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(A) Let the initial velocity be $u$ and uniform acceleration be $a$.For the first interval of $t = 4 \,sec$,the distance $s_1 = 24 \,m$:$s = ut + \frac{1}{2}at^2$$24 = u(4) + \frac{1}{2}a(4)^2$$24 = 4u + 8a$Dividing by $4$,we get: $6 = u + 2a$ ... $(i)$For the total time $t = 8 \,sec$,the total distance $s_{total} = 24 + 64 = 88 \,m$:$88 = u(8) + \frac{1}{2}a(8)^2$$88 = 8u + 32a$Dividing by $8$,we get: $11 = u + 4a$ ... $(ii)$Subtracting equation $(i)$ from $(ii)$:$(u + 4a) - (u + 2a) = 11 - 6$$2a = 5 \implies a = 2.5 \,m/s^2$Substituting $a = 2.5$ in equation $(i)$:$6 = u + 2(2.5)$$6 = u + 5$$u = 1 \,m/s$.

$A$ particle moving with a uniform acceleration travels $24 \,m$ and $64 \,m$ in the first two consecutive intervals of $4 \,sec$ each. Its initial velocity is............$m/sec$.

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