r(t) = (2t^2,4t^2-t^3)
v(t) = dr/dt = (4t,8t-3t^2)
a(t) = dv/dt = (4,8-6t)
at t=1,
a(1) = (4,2)
|a(1)| = √(4^2+2^2) = 3√2
Hmm. I suspect a typo or a wrong answer key.
Find the magnitude of a particle's acceleration vector at t = 1 for a particle moving with position vector r(t)=(2t^2,4t^2-t^3)
a) 4
b) 25
c) 5
d) 2√5
2 answers
A typo indeed.
|a(1)| = √(4^2+2^2) = √20 = 2√5
I forgot to square the 2!
|a(1)| = √(4^2+2^2) = √20 = 2√5
I forgot to square the 2!
2 answers