(b) $x = 4(t - 2) + a(t - 2)^{2}$ At $t = 0,\ x = - 8 + 4a = 4a - 8$ $v = \frac{dx}{dt} = 4 + 2a(t - 2)$ At $t = 0,\ v = 4 - 4a = 4(1 - a)$ But acceleration, $a = \frac{d^{2}x}{dt^{2}} = 2a$
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