Architecture
건축 Architecture


Time for encounter of two objects moving with different accelerations

작성자 Uploader : deyoon 작성일 Upload Date: 2016-04-15변경일 Update Date: 2016-04-15조회수 View : 230

Travel distance of an object with its initial velocity and constant acceleration being V1 and a1, from the initial position C;
S(t) = Vt+1/2at^2 + C

thus, positions S1 and S2 of two objects(P1 and P2) with respect to time are:
S1(t) = V1*t+1/2*a1*t^2+C1
S2(t) = V2*t+1/2*a2*t^2+C2

At the position where two objects encounter, S1(t)=S2(t)
i.e. from the quadratic equation : 1/2(a1-a2)*t^2 + (V1-V2)*t + (C1-C2) = 0  

We get the t as:  
t = (-(V1-V2) ± ((V1-V2)^2 - 4*1/2*(a1-a2)*(C1-C2))^(1/2))/(2*(1/2)*(a1-a2))  
t = (1/(a1-a2))*(-(V1-V2) ± ((V1-V2)^2 - 2*(a1-a2)*(C1-C2))^(1/2))  

*Note: set (a1-a2) to be positive and take the positive value of t.



*** 참고문헌[References] ***

No reference
t = (1/(a1-a2))*(-(V1-V2) + ((V1-V2)^2 - 2*(a1-a2)*(C1-C2))^(1/2))
변수명 Variable 변수값 Value 변 수 설 명 Description of the variable




★ 로그인 후 수식작성 및 즐겨찾기에 추가할 수 있습니다.
★ To make new formula or to add this formula in your bookmark, log on please.


코멘트

댓글 입력