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Can you answer this calculus question?

A particle moves along the graph of y=cos x so that its x-component of acceleration is always 2. At time t=0, the particle is at the point (pi, -1) and the velocity of the particle is <0, 0>.
a) Find the position vector of the particle.
b) Find the speed of the particle when it is at the point (4, cos 4).
Here's my work for part a):
s(t)=<f(t), cos(f(t))>
v(t)=<f'(t), -sin(f(t))f'(t)>
a(t)=<f"(t), -sin(f(t))f"(t)-cos(f(t))f"(t)>
=<2, -sin(f(t))f"(t)-cos(f(t))f"(t)>
I know that f"(t)=2 and f'(t)=2t+C but what's C? I'm stucked. Please help me step by step. Thanks.
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1 Answer

       v(t)=<f'(t), -sin(f(t))f'(t)> you were given that at t=0 v(0)=<0,0>, and f'(t)=2t+C so C=0 and f'(t)=2t.
So now f(t)=t2+b and you calculate b from the other initial condition at t=0 x=pi and y=-1