A particle moves along the x-axis so that at any time t, measured in seconds, its position is given by s(t) = 5cos(t) − sin(3t), measured in feet. What is the acceleration of the particle at time t = π seconds?
Your turn
To find the acceleration of the particle at time t = π seconds, we need to find the second derivative of the position function s(t). The second derivative will give us the acceleration function.
Let's start by finding the first derivative of s(t) with respect to t:
s'(t) = d/dt (5cos(t) − sin(3t))
To find the derivative of each term, we can use the chain rule. The derivative of cos(t) is -sin(t), and the derivative of sin(3t) is 3cos(3t). Applying the chain rule, we have:
s'(t) = -5sin(t) - 3cos(3t)
Now, let's find the second derivative, denoted as s''(t), by taking the derivative of s'(t) with respect to t:
s''(t) = d/dt (-5sin(t) - 3cos(3t))
Again, using the chain rule, the derivative of -5sin(t) is -5cos(t), and the derivative of -3cos(3t) is 9sin(3t).
So, s''(t) = -5cos(t) - 9sin(3t)
Now, to find the acceleration of the particle at t = π seconds, we substitute π into the acceleration function:
a(π) = -5cos(π) - 9sin(3π)
Using the trigonometric identities cos(π) = -1 and sin(3π) = 0, we have:
a(π) = -5(-1) - 9(0)
a(π) = 5
Therefore, the acceleration of the particle at time t = π seconds is 5 feet per square second.