直線運動選擇題 求好心人士幫忙?

a = dV / dt = 5
因此為等加速度

Vo = V( t = 0 ) = - 10 + 5*0 = - 10
故初速為 - 10

X
= Vo t + (1/2) a t²
= - 10t + (1/2) 5 t²
= - 10t + 2.5t²

所以位移函數為 :
X(t)
= - 10t + 2.5t²
= 2.5( t - 2 )² - 10
在 t = 2 時, 有最小值為 - 10

X(0) = 0
X(1) = - 10 + 2.5 = - 7.5
X(2) = - 10 (最小值)
X(3) = 2.5( 3 - 2 )² - 10 = 2.5 - 10 = - 7.5
X(4) = 2.5( 4 - 2 )² - 10 = 10 - 10 = 0
X(5) = 2.5( 5 - 2 )² - 10 = 2.5*9 - 10 = 12.5

0 < t < 2 時, 此物向西運動
t = 2 時, 開始折返
t > 2 時, 此物向東運動

Ans : D