UVA10071 Back to High School Physics:題解
題目連結:https://www.luogu.org/problemnew/show/UVA10071
題意簡敘:
粒子從0速度提速到t時速度為v,求出2*t時所運動的距離
分析:
這道題是一道物理題 ,首先我們知道粒子最初的速度是0,又已知其在t時間時速度為v,由
a=△v/△t,
所以
a=(v−0)/(t−0),
即
at=v
在這裡可以用公式:
距離S=初速度V*時間t+1/2at^2
求解。
即
S=0×2t+½a(2t)2
化簡得
S=2at2
結合剛才推出的 at=v
∴ S=2×v×t
∵ v,t均已知
∴問題得解。
程式碼:
#include<cstdio>
using namespace std;
int main()
{
int v,t;
while(scanf("%d%d",&v,&t)==2)
printf("%d\n",2*v*t);
return 0;
}
撒花~
求贊QAQ
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