(C) Given the distance function $s = t^2 - 2t + 5$. Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = \f

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(C) Given the distance function $s = t^2 - 2t + 5$. Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = \frac{d}{dt}(t^2 - 2t + 5) = 2t - 2$. Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(2t - 2) = 2$. Thus, the acceleration of the particle is $2$ units.

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If the distance $s$ travelled by a particle in time $t$ is given by $s=t^2-2t+5$, then its acceleration is

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If the distance $s$ travelled by a particle in time $t$ is given by $s=t^2-2t+5$, then its acceleration is

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