^{2}+b and you calculate b from the other initial condition at t=0 x=pi and y=-1

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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Sun,

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)=t^{2}+b and you calculate b from the other initial condition at t=0 x=pi and y=-1

Jim

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