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(B) The given equation of motion is $s = at^2 + bt + c$. Displacement after $1 	ext{ s}$ is $20 	ext{ m}$,so $s(1) = a(1)^2 + b(1) + c = 20 \implies

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$a + c = b$

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(B) The given equation of motion is $s = at^2 + bt + c$. Displacement after $1 	ext{ s}$ is $20 	ext{ m}$,so $s(1) = a(1)^2 + b(1) + c = 20 \implies a + b + c = 20 \dots (i)$. Velocity $v = \frac{ds}{dt} = 2at + b$. Velocity after $2 	ext{ s}$ is $30 	ext{ m/s}$,so $v(2) = 2a(2) + b = 30 \implies 4a + b = 30 \dots (ii)$. Acceleration $a_{acc} = \frac{dv}{dt} = 2a$. Given acceleration is $10 	ext{ m/s}^2$,so $2a = 10 \implies a = 5$. Substituting $a = 5$ into equation $(ii)$: $4(5) + b = 30 \implies 20 + b = 30 \implies b = 10$. Substituting $a = 5$ and $b = 10$ into equation $(i)$: $5 + 10 + c = 20 \implies 15 + c = 20 \implies c = 5$. Now,checking the options: $a + c = 5 + 5 = 10$ and $b = 10$. Therefore,$a + c = b$.

The equation of motion of a particle is $s = at^2 + bt + c$. If the displacement after $1 \text{ s}$ is $20 \text{ m}$,velocity after $2 \text{ s}$ is $30 \text{ m/s}$,and the acceleration is $10 \text{ m/s}^2$,then:

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