F.6 數學定積分問題6

2009-10-12 5:09 am

A和B兩車作直路上10秒的衝刺比試.這兩車於時間t秒的速度(單位:m/s)分別為

car A : v(t)=t(10-t)

car B : v(t)=(1/5)(t^2)(10-t)

(a)求這兩車所走的距離

(b)求這兩車的最高速度

回答 (1)

Prof. Physics

2009-10-12 5:19 am

✔ 最佳答案

a. A所走的距離

= S (t = 0 → 10) t(10 - t) dt

= S (t = 0 → 10) (-t2 + 10t) dt

= [-t3/3 + 5t2] (t = 0 → 10)

= 500/3 m

B所走的距離

= S (t = 0 → 10) 1/5 t2(10 - t) dt

= 1/5 S (t = 0 → 10) (-t3 + 10t2) dt

= 1/5 [-t4/4 + 10t3/3](t = 0 → 10)

= 500/3 m

車A，

dv(t)/dt = -t + 10 - t = 10 - 2t

d2v(t)/dt2 = -2 < 0

設dv(t)/dt = 0, t = 5

所以，t = 5，車A是最高速度 = 5(10 - 5) = 25 m/s

車B，

dv(t)/dt = 1/5 [2t(10 - t) - t2] = 1/5 (-3t2 + 20t)

d2v(t)/dt2 = 1/5 (-6t + 20)

dv(t)/dt = 0, t = 0 或 20/3

當t = 0，d2v(t)/dt2 = 4 > 0

當t = 20/3，d2v(t)/dt2 = -4 < 0

所以，當t = 20/3，車B是最高速度 = 1/5 (20/3)2(10 - 20/3) = 800/27 m/s

參考: Physics king

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