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(B) Let the time taken for acceleration be $t_1$ and the time taken for deceleration be $t_2$. Given,total time $T = t_1 + t_2 = 4\, s$. The velocity-

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$16\, m$

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(B) Let the time taken for acceleration be $t_1$ and the time taken for deceleration be $t_2$. Given,total time $T = t_1 + t_2 = 4\, s$. The velocity-time graph for this motion is a triangle with base $T = 4\, s$ and height $v_{max} = 8\, m s^{-1}$. The distance travelled is equal to the area under the velocity-time graph. Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. Area $= \frac{1}{2} 	imes 4\, s 	imes 8\, m s^{-1} = 16\, m$. Thus,the total distance travelled is $16\, m$.

$A$ particle accelerates from rest at a constant rate for some time and attains a constant velocity of $8\, m s^{-1}$. Afterwards,it decelerates at a constant rate and comes to rest. If the total time taken is $4\, s$,the distance travelled is:

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