(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 =

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(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 $b = -33$.

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

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$A$ particle starts moving from rest from a fixed point in a fixed direction. The distance $s$ from the fixed point at a time $t$ is given by $s = t^{2} + at - b + 17$,where $a$ and $b$ are real numbers. If the particle comes to rest after $5 \ s$ at a distance of $s = 25$ units from the fixed point,then the values of $a$ and $b$ are respectively:

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A
$a = -10, b = -33$
B
$a = -10, b = -30$
C
$a = -8, b = 33$
D
$a = -10, b = 33$

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