看板Examination
: 這題一直不知道怎麼算..
: When an automobiles accelerated forwarding from stand still to high speed
: on a flat ground , the acceleration at time t=4 sec is about ?
: V-T : http://ppt.cc/jziw
: Thanks a lot !
觀察圖中, a(t)為線性函數, 故假設a(t)=bt+c 其中b,c為常數
又v(t)=∫a(t)dt =∫(bt+c)dt =0.5bt^2+ct+d 其中d為常數
v(0)=0帶入v(t),得d=0
故 v(t)=0.5bt^2+ct 且 a(t)=bt+c
v(10)=50,a(0)=10帶入 => 50=50b+10c c=10 => 解得b=-1,c=10
=>v(t)=-0.5t^2+10t a(t)=-t+10, 解得a(4)=6 m/s2
......原本是這樣算,可是原po說答案a(4)=4
所以又再觀察一下圖,覺得a(t)應該通過原點且跟t成正比,就假設題目a0=10打錯,改成a0=0
=>v(10)=50,a(0)=0帶入 => b=1 c=0
=>v(t)=0.5t^2 a(t)=t, 解得a(4)=4 m/s2
當然這樣v(t)圖又錯了,a(t)斜率為正的話v(t)圖應該上凹才對
但至少答案對了(?
--
◆ From: 218.35.161.42
※ 批踢踢實業坊(ptt.cc)※ 文章網址: https://www.ptt.cc/bbs/Examination/M.1384066844.A.952.html
→ hoks:...a(0)=0的話那張圖從頭到尾都有問題 11/10 15:30
→ hoks:他那張圖的a(0)的線就應該貼著t軸跑才對 11/10 15:31
→ hoks:而且方程式裡面會有2次微分的值存在 會有反曲點 11/10 15:34