A bird flies with a velocity (t-2) text{ m/s} | vedclass

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(B) The velocity of the bird is given by $v(t) = t - 2$. Since distance is the integral of the magnitude of velocity, we calculate: $	ext{Distance} =

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$4$

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(B) The velocity of the bird is given by $v(t) = t - 2$. Since distance is the integral of the magnitude of velocity, we calculate: $	ext{Distance} = \int_{0}^{4} |v(t)| 	ext{ dt} = \int_{0}^{4} |t - 2| 	ext{ dt}$. We split the integral at $t = 2 	ext{ s}$ where the velocity changes sign: $	ext{Distance} = \int_{0}^{2} -(t - 2) 	ext{ dt} + \int_{2}^{4} (t - 2) 	ext{ dt}$. Evaluating the first part: $\int_{0}^{2} (2 - t) 	ext{ dt} = [2t - \frac{t^2}{2}]_{0}^{2} = (4 - 2) - 0 = 2 	ext{ m}$. Evaluating the second part: $\int_{2}^{4} (t - 2) 	ext{ dt} = [\frac{t^2}{2} - 2t]_{2}^{4} = (8 - 8) - (2 - 4) = 0 - (-2) = 2 	ext{ m}$. Total distance $= 2 	ext{ m} + 2 	ext{ m} = 4 	ext{ m}$.

$A$ bird flies with a velocity $(t-2) \text{ m/s}$ along a straight line, where $t$ is the time in seconds. The distance covered by it in a time of $4 \text{ s}$ is: (in $\text{ m}$)

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