物理 - 考試

※ 引述《yyyyu (mm)》之銘言：
: 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(?


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