The position of a cat running from a dog down a dark alley i
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The position of a cat running from a dog down a dark alley is given by the values of the table.
t(seconds)012345s(feet)020306990107
Draw the graph of the function for yourself and estimate the instantaneous velocity of the cat (ft/sec) when t=2
t(seconds)012345s(feet)020306990107
Draw the graph of the function for yourself and estimate the instantaneous velocity of the cat (ft/sec) when t=2
Answer:The instantaneous velocity of the cat is 20 ft/sec when t=2.
【解析】按照表格数据,很明显猫咪是在做变速运动,小狗所在的位置为起点,昏暗的巷子为跑道.则按照物理学公式,t=1~5时,对应的加速度分别用a1~a5表示,对应的瞬时速度分别用v1~v5表示,位移s分别用s1~s5表示;t=0时,初速度用V0表示;此外,Δt表示时间差.
【解】依题意,当t=0时,s=0,所以有初速度v0=0;
t=1时,s1=v0×Δt+(1/2)×a1×Δt=0×(1-0)+(1/2)×a1×(1-0)=(1/2)×a1=20,解得a1=40(ft/sec)
所以v1=v0+a1×Δt=0+40×(1-0)=40(ft/sec);
t=2时,s2=v1×Δt+(1/2)×a2×Δt=40×(2-1)+(1/2)×a2×(2-1)=40+(1/2)×a2=30,解得a2=-10(ft/sec)
所以v2=v1+a2×Δt=40+(-10)×(2-1)=20(ft/sec)
答:t=2sec时,对应的瞬时速度为20 ft/sec.
赠人玫瑰,手留余香!——福来美Flower Magic
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【解析】按照表格数据,很明显猫咪是在做变速运动,小狗所在的位置为起点,昏暗的巷子为跑道.则按照物理学公式,t=1~5时,对应的加速度分别用a1~a5表示,对应的瞬时速度分别用v1~v5表示,位移s分别用s1~s5表示;t=0时,初速度用V0表示;此外,Δt表示时间差.
【解】依题意,当t=0时,s=0,所以有初速度v0=0;
t=1时,s1=v0×Δt+(1/2)×a1×Δt=0×(1-0)+(1/2)×a1×(1-0)=(1/2)×a1=20,解得a1=40(ft/sec)
所以v1=v0+a1×Δt=0+40×(1-0)=40(ft/sec);
t=2时,s2=v1×Δt+(1/2)×a2×Δt=40×(2-1)+(1/2)×a2×(2-1)=40+(1/2)×a2=30,解得a2=-10(ft/sec)
所以v2=v1+a2×Δt=40+(-10)×(2-1)=20(ft/sec)
答:t=2sec时,对应的瞬时速度为20 ft/sec.
赠人玫瑰,手留余香!——福来美Flower Magic
楼主以后有需要解答的题目,可以直接百度信息给我.
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