A particle moving with a uniform acceleration | vedclass

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Question

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

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$1$

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