A particle moves along the x-axis so that at any time t, measured in seconds, its position is given by s(t)equals = 5cos(t) minus- sin(3t), measured in feet. What is the acceleration of the particle at time t =equals π seconds?
Is the answer 5?
s(t)equals = 5cos(t) minus- sin(3t) ???
why all the words? Just write
s(t) = 5cos(t) - sin(3t)
v(t) = -5sin(t) - 3cos(3t)
a(t) = -5cos(t) + 9sin(3t)
a(π) = -5(-1) + 9(0) = 5
you are correct.
To find the acceleration of the particle at time t = π seconds, we need to find the second derivative of the position function.
Given that the position function is s(t) = 5cos(t) - sin(3t), we can find the velocity function by taking the derivative of the position function:
v(t) = d/dt (5cos(t) - sin(3t))
Differentiating each term separately, we get:
v(t) = -5sin(t) - 3cos(3t)
Now, to find the acceleration function, we need to take the derivative of the velocity function:
a(t) = d/dt (-5sin(t) - 3cos(3t))
Differentiating each term separately, we get:
a(t) = -5cos(t) + 9sin(3t)
Now we can substitute t = π to find the acceleration at that time:
a(π) = -5cos(π) + 9sin(3π)
Since cos(π) = -1 and sin(3π) = 0, the equation simplifies to:
a(π) = -5(-1) + 9(0)
a(π) = 5
Therefore, the acceleration of the particle at time t = π seconds is 5 feet per second squared.
To find the acceleration of the particle at time t = π seconds, we need to take the second derivative of the position function, s(t).
Step 1: Find the first derivative of s(t):
The first derivative of a function represents the velocity, so let's differentiate s(t) with respect to t.
s'(t) = d/dt [5cos(t) - sin(3t)]
To find the derivative, we apply the chain rule and the derivative of each term:
s'(t) = -5sin(t) - 3cos(3t)
Step 2: Find the second derivative of s(t):
The second derivative of a function represents the acceleration, so let's differentiate s'(t) with respect to t.
s''(t) = d/dt [-5sin(t) - 3cos(3t)]
Again, we apply the chain rule and the derivative of each term:
s''(t) = -5cos(t) + 9sin(3t)
Step 3: Evaluate the acceleration at t = π seconds.
To find the acceleration at t = π seconds, substitute π into the equation for s''(t):
s''(π) = -5cos(π) + 9sin(3π)
Remember that cos(π) = -1 and sin(3π) = 0, so:
s''(π) = -5(-1) + 9(0)
s''(π) = 5
Therefore, the acceleration of the particle at time t = π seconds is 5 feet/second².