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A particle is moving along $y$-axis. The position $y$ of particle varies with time $t$ as $y=\left(t^{3}-t^{2}+1\right) \mathrm{m}$. The displacement of particle from $t=0$ to $t=2 \mathrm{~s}$ is[1](a) $1 \mathrm{~m}$(b) $4 \mathrm{~m}$(c) $-1 \mathrm{~m}$(d) $-4 m$

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## Solution 1

#### Solution By Steps***Step 1: Calculate position at $t=0$***Substitute $t=0$ into the position function $y = t^3 - t^2 + 1$:$$ y(0) = 0^3 - 0^2 + 1 = 1 ext{ m} $$***Step 2: Calculate position at $t=2$***Substitute $t=2$ into the position function $y = t^3 - t^2 + 1$:$$ y(2) = 2^3 - 2^2 + 1 = 8 - 4 + 1 = 5 ext{ m} $$***Step 3: Calculate displacement***Displacement is the difference in position between $t=2$ and $t=0$:$$ ext{Displacement} = y(2) - y(0) = 5 - 1 = 4 ext{ m} $$#### Final AnswerThe displacement of the particle from $t=0$ to $t=2 ext{ s}$ is $4 ext{ m}$.The correct option is (b) $4 ext{ m}$.

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