Question
A particle is moving with the given data. Find the position of the particle.
a(t)= (t^2 - 5t +7) , s(0)=0 , s(1)=20
Thanks!
a(t)= (t^2 - 5t +7) , s(0)=0 , s(1)=20
Thanks!
Answers
Steve
a = t^2-5t+7
v = 1/3 t^3 - 5/2 t^2 + 7t + C1
s = 1/12 t^4 - 5/6 t^3 + 7/2 t^2 + C1*t + C2
plugging s(0) and s(1), we have
C1 = 0
1/12 - 5/6 + 7/2 + 0 + C2 = 20
C2 = 69/4
so,
s(t) = 1/12 t^4 - 5/6 t^3 + 7/2 t^2 + 69/4
v = 1/3 t^3 - 5/2 t^2 + 7t + C1
s = 1/12 t^4 - 5/6 t^3 + 7/2 t^2 + C1*t + C2
plugging s(0) and s(1), we have
C1 = 0
1/12 - 5/6 + 7/2 + 0 + C2 = 20
C2 = 69/4
so,
s(t) = 1/12 t^4 - 5/6 t^3 + 7/2 t^2 + 69/4
Anonymous
When I put your answer into WebAssign it marked as wrong so are you sure this is the right answer?
Steve
well, pretty sure. Do you see any errors in my work?
wolframalpha agrees with me.
http://www.wolframalpha.com/input/?i=solve+y%22+%3D+x^2-5x%2B7+where+y%280%29%3D0+and+y%281%29%3D20
wolframalpha agrees with me.
http://www.wolframalpha.com/input/?i=solve+y%22+%3D+x^2-5x%2B7+where+y%280%29%3D0+and+y%281%29%3D20
Steve
actually, I do see a mistake.
C2 = 0
C1 = 69/4
s(t) = 1/12 t^4 - 5/6 t^3 + 7/2 t^2 + 69/4 t
C2 = 0
C1 = 69/4
s(t) = 1/12 t^4 - 5/6 t^3 + 7/2 t^2 + 69/4 t