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(C) The change in velocity $\Delta v$ is equal to the area under the acceleration-time graph.For the particle to acquire its initial velocity,the tota

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$8$

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(C) The change in velocity $\Delta v$ is equal to the area under the acceleration-time graph.For the particle to acquire its initial velocity,the total change in velocity must be zero,i.e.,$\Delta v = 0$.This implies that the net area under the $a-t$ graph must be zero.Let the area above the time axis be $A_1$ and the area below the time axis be $A_2$.$A_1 = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4 	imes 10 = 20 \ m/s$.Let the slope of the line be $m = \frac{0 - 10}{4 - 0} = -2.5 \ m/s^3$.The equation of the line is $a = -2.5t + 10$.At time $t_0$,the acceleration is $a(t_0) = -2.5t_0 + 10$.The area $A_2$ is a triangle with base $(t_0 - 4)$ and height $|a(t_0)| = | -2.5t_0 + 10 | = 2.5t_0 - 10$.Since $A_1 = A_2$,we have $20 = \frac{1}{2} 	imes (t_0 - 4) 	imes (2.5t_0 - 10)$.$40 = (t_0 - 4) 	imes 2.5(t_0 - 4)$.$40 = 2.5(t_0 - 4)^2$.$(t_0 - 4)^2 = \frac{40}{2.5} = 16$.$t_0 - 4 = 4$.$t_0 = 8 \ s$.

The acceleration-time graph of a particle moving along a straight line is shown. At what time does the particle acquire its initial velocity? (in $s$)

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