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
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!
4 answers
When I put your answer into WebAssign it marked as wrong so are you sure this is the right answer?
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
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