A bird flies for 4 , s with a velocity of |t | vedclass

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(B) The velocity of the bird is given by $v(t) = |t - 2| \, m/s$.To find the distance covered in $4 \, s$,we integrate the speed with respect to time

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$4$

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(B) The velocity of the bird is given by $v(t) = |t - 2| \, m/s$.To find the distance covered in $4 \, s$,we integrate the speed with respect to time from $t = 0$ to $t = 4 \, s$:$S = \int_{0}^{4} |t - 2| \, dt$We can split the integral at $t = 2$ because the expression inside the absolute value changes sign:$S = \int_{0}^{2} -(t - 2) \, dt + \int_{2}^{4} (t - 2) \, dt$$S = \int_{0}^{2} (2 - t) \, dt + \int_{2}^{4} (t - 2) \, dt$$S = [2t - \frac{t^2}{2}]_{0}^{2} + [\frac{t^2}{2} - 2t]_{2}^{4}$$S = (4 - 2) - (0) + [(8 - 8) - (2 - 4)]$$S = 2 + [0 - (-2)] = 2 + 2 = 4 \, m$.Alternatively,the distance is the area under the velocity-time graph,which consists of two right-angled triangles,each with base $2 \, s$ and height $2 \, m/s$.$S = 2 	imes (\frac{1}{2} 	imes 2 	imes 2) = 4 \, m$.

$A$ bird flies for $4 \, s$ with a velocity of $|t - 2| \, m/s$ in a straight line,where $t$ is time in seconds. It covers a distance of ......... $m$.

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