Answer:
The acceleration of the particle at time t = 0 seconds is:
4 feet per square second.
i.e. 4 ft/s²
Step-by-step explanation:
We are given a position function in terms of the time t as:
\(s(t)=4\sin (t)-\cos (2t)\)
Now, we are asked to find the acceeleration of the particle at time t = 0 seconds.
We know that the acceleration of a particle is given by:
\(a=\dfrac{d}{dt}v\)
where v is the velocity of the particle which is calculated by:
\(v=\dfrac{d}{dt}s\)
Hence, we get:
\(a=\dfrac{d}{dt}(\dfrac{d}{dt}s)\\\\i.e.\\\\a=\dfrac{d^2}{dt^2}s\\\\i.e.\\\\a=s''(t)\)
i.e. the acceleration of the particle is the double derivative of the position.
\(s'(t)=4\cos (t)+2\sin (2t)\)
and
\(s''(t)=-4\sin (t)+4\cos (2t)\)
i.e.
\(a(t)=-4\sin (t)+4\cos (2t)\\\\i.e.\ at\ t=0\ we\ have:\\\\a(0)=-4\sin 0+4\cos 0\\\\i.e.\\\\a(0)=4\)