Question

A particle moves along the x-axis so that at any time t, measured in seconds, its position is given by s(t) = 4sin(t) - cos(2t), measured in feet. What is the acceleration of the particle at time t = 0 seconds?

Answer 1

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.

$a(t)=-4\sin (t)+4\cos (2t)\\\\i.e.\ at\ t=0\ we\ have:\\\\a(0)=-4\sin 0+4\cos 0\\\\i.e.\\\\a(0)=4$