A particle starts moving from rest from a fix | vedclass

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Question

(A) Given the distance equation: $s = t^{2} + at - b + 17$. Since the particle starts from rest, the velocity $v = \frac{ds}{dt}$ must be zero at $t =

Vedclass · Accepted Answer

$a = -10, b = -33$

Vedclass · Answer

(A) Given the distance equation: $s = t^{2} + at - b + 17$. Since the particle starts from rest, the velocity $v = \frac{ds}{dt}$ must be zero at $t = 5 \ s$. Calculating the velocity: $v = \frac{ds}{dt} = 2t + a$. Setting $v = 0$ at $t = 5$: $2(5) + a = 0 \implies 10 + a = 0 \implies a = -10$. Now, given that at $t = 5 \ s$, the distance $s = 25$: $25 = (5)^{2} + a(5) - b + 17$. Substituting $a = -10$: $25 = 25 + (-10)(5) - b + 17$. $25 = 25 - 50 - b + 17$. $25 = -8 - b$. $b = -8 - 25 = -33$. Thus, the values are $a = -10$ and $b = -33$.

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